Trigonometry Teacher’s Edition - Solution Key CK-12 Foundation December 9, 2009 CK-12 Foundation is a non-proﬁt organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the “FlexBook,” CK-12 intends to pioneer the generation and distribution of high quality educational content that will serve both as core text as well as provide an adaptive environment for learning. Copyright ©2009 CK-12 Foundation This work is licensed under the Creative Commons Attribution-Share Alike 3.0 United States License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/us/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA. Contents 1 TE Trigonometry and Right Angles - Solution Key 1.1 Trigonometry and Right Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 TE Circular Functions - Solution Key 2.1 Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 TE Inverse Functions and Trigonometric Equations - Solution Key 4.1 49 49 103 137 Triangles and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 6 TE Polar Equations and Complex Numbers - Solution Key 6.1 33 Inverse Functions and Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 103 5 TE Triangles and Vectors - Solution Key 5.1 5 33 3 TE Trigonometric Identities - Solution Key 3.1 5 237 Polar Equations and Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 3 4 Chapter 1 TE Trigonometry and Right Angles Solution Key 1.1 Trigonometry and Right Angles Basic Functions Review Exercises: 1. a) This relation is not a function. The x−value of 1 is paired with two y−values: 5 and 7. b) This relation is a function. Any vertical line will cross the graph of y = 3 − x only once. Each x−value is paired with one and only one y−value. x y −3 6 −2 5 −1 4 0 3 1 2 2 1 c) This relation is not a function. Any vertical line will cross the graph more than once. 5 3 0 2. a) distance = rate · time d = 95t b) This situation is direct variation because as the time increases the distance increases at the same rate. c) d = 95t → general equation d = 95 miles/hr (3hr) → given t = 3 hours d = 285 miles. 3. a) y = mx + b. y represents the cost of beginning the business; m represents the cost of each wooden frame(x) and b represents the initial output of money (0, 100). y = 2x + 100 c(x) = 2x + 100 → y = 2x + 100 written as a function. b) y = mx + b. y represents the revenue; m represents the selling price of each picture frame and b represents any other revenue which in this case is zero. y = 10x R(x) = 10x → y = 10x written as a function. c) P (x) = R(x) − C(x) → The proﬁt P (x) is the diﬀerence between the revenueR(x) and the cost C(x) P (x) = 10x − (2x + 100) P (x) = 10x − 2x − 100 P (x) = 8x − 100 4. a) The function deﬁned by the equation f (x) = x2 − x − 3 is of the general form of a quadratic function. b) The domain of the function is: {xIxεR} The range of the function is: {yIy ≥ −3.25, yεR} c) Using the TI – 83 to graph f (x) = x2 − x − 3 The coordinates of the vertex and the x−intercepts can be determined by using the 2nd Trace function: 6 Vertex x−intercepts The vertex is (5.0, −3.25) and the x−intercepts are (−1.3, 0) and (2.3, 0). 5. a) Using the TI – 83 to graph y = x−2 x+3 : The asymptotes are x = −3 and y = 1 6. a) c = p1 (500) → the cost per person (c) of renting a party room varies inversely with the number of people who attend and the initial cost of renting the room (500) b) 1 (500) → given p = 32 p 1 c= (500) 32 c = $15.63 c= 7. a) Using the TI – 83 to graph: y = x3 y = x3 + x y = x3 + 2x 7 The equations with positive coeﬀicients look more and more like y = x3 , as the coeﬀicient gets larger. y = x3 y = x3 − x y = x3 − 2x The equations with negative coeﬀicients have local maximums and minimums. Decreasing the coeﬀicient increases the size of the” hill” and the “valley.” 8. a) Using the TI – 83 to graph the function p(x) = −.5x2 + 90x − 200: The number of units that must be sold to attain the maximum proﬁt is the vertex of the parabola. Use 2nd Trace 8 The maximum proﬁt is $3850 with 90 units being sold. The x−intercepts are: and The x−intercepts represent the break-even points of the company. The company must sell at least 2.25 units to cover any initial costs but when 177.7 units are sold, it no longer makes a proﬁt. 9. a) Using the TI – 83 to create a scatter plot of the given data: b) The period is twelve months. c) The number of daylight hours in other areas would not show as much variance, so the amplitude of the graph would be smaller. Angles in Triangles Review Exercises: 9 1. △ABC is isosceles. An isosceles triangle has two sides equal in length. Therefore AC is either 5 inches in length or 7 inches in length. 2. An obtuse triangle is one that has one angle that measures greater than 90◦ . A right triangle is one that has one angle that measures 90◦ . The sum of the angles of a triangle is 180◦ . Therefore a right triangle has one angle of 90◦ and two acute angles. A right triangle cannot be an obtuse triangle. 3. In any triangle, the sum of the three angles is 180◦ . ∠1 + ∠2 + ∠3 = 180◦ 48◦ + 28◦ + ∠ = 180◦ 76◦ + ∠3 = 180◦ ∠3 = 180◦ − 76◦ ∠3 = 104◦ 4. a) ∠1 + ∠2 + ∠3 = 180◦ 90◦ + ∠2 + ∠3 = 180◦ ∠2 + ∠3 = 180◦ − 90◦ ∠2 + ∠3 = 90◦ Complementary Angles are two angles whose sum equals 90◦ . Therefore, the two acute angles of a right triangle are complementary angles. b) ∠1 = 90◦ → given ∠2 + ∠3 = 90◦ 23◦ + ∠3 = 90◦ ∠3 = 90◦ − 23◦ ∠3 = 67◦ 5. Let x represent ∠D. ∠O = 2x since the measure of ∠O is twice the measure of ∠D and ∠G = 3x since the measure of ∠G is three times the measure of ∠D. 10 Therefore:x + 2x + 3x = 180◦ 6x = 180◦ 6x 180◦ = 6 6 x = 30◦ ∠D = x = 30◦ ∠O = 2x = 60◦ ∠G = 3x = 90◦ 6. BC AC = EF DF 8 10 = 6 DF 8DF = 60 8DF 60 = 8 8 8 = 7.5 DF 7. If two triangles are similar, then the corresponding angles are congruent. Therefore, ∠B = ∠E. In △ABC, ∠A = 30◦ and ∠C = 20◦ . ∠B = 180◦ − (30◦ + 20◦ ) ∠B = 130◦ ∴ ∠E = 130◦ 8. a) AT OG = CT DG 8 6 = 12 8 3 2 ̸= 3 4 △ACT and △DOG are not similar. b) AB BC AC = = DE EF DF 12 5 13 = = 6.5 6 2.5 2=2=2 △ABC and △DEF are similar. 11 9. 20 24 = x 100 24x = 2000 24x 2000 = 24 24 1 x = 83 feet 3 The height of the building is 83 13 feet. 10. The answers to this question will vary. However, the answer should include the fact that corresponding sides of similar triangles are proportional and that corresponding angles are congruent. Measuring Rotation Review Exercises: 1. a) This angle is less than 90◦ and is an acute angle. b) This angle is a rotation of 180◦ and is a straight angle. 2. a) The measure of this angle is greater than 90◦ but less than 180◦ . Since the terminal arm of the angle is less than half way between 90◦ and 180◦ , the approximate measure of the angle is 120◦ . A protractor could be used to determine the exact measure of the angle. 3. a) 85.5◦ expressed in degrees, minutes and seconds would be 85◦ 30′ 50 x = 100 60 100x = 3000 100x 3000 = 100 100 x = 30 b) 12.15◦ expressed in degrees, minutes and seconds would be 12◦ 9′ . 12 15 x = 100 60 100x = 900 100x 900 = 100 100 x=9 c) 114.96◦ expressed in degrees, minutes and seconds would be 114◦ 57′ 3.6” 96 x = 100 60 100x = 5760 100x 5760 = 100 100 x = 67.6 0.6 s = 60 360 60s = 216 60s 216 = 60 60 s = 3.6 4. a) 10 25 + 60 360 60 25 ◦ 54 + + 360 360 85 ◦ 54 + 360 54◦ + 0.236 54◦ + ≈ 54.236◦ b) 40 5 + 60 360 240 5 ◦ 17 + + 360 360 245 ◦ 17 + 360 17◦ + 0.681 17◦ + ≈ 17.681◦ 5. a) 13 The angle between the hands of the clock at 6:00 is 180◦ . b) The angle between the hands of the clock at 3:00 is 90◦ . c) The angle between the hands of the clock at 1:00 is 30◦ . 6. Between 12:00 and 1:00 o’clock, the arms of the clock rotate through an angle of 360◦ . 7. 1 πd 4 1 = (π)(200 m) 4 = 50(π)meters Cinner track = Cinner track Cinner track 14 Couter wheel = πd Couter wheel = (π)(0.6m) Couter wheel = 0.6(π)meters 1 πd 4 1 = ((π)((204 m) 4 = 51(π) meters Couter track = Couter track Couter track Cinner wheel = πd Coinner wheel = (π)(0.6 m) Cinner wheel = 0.6(π) meters Couter track − Cinner track = 1π and 1π 5 ≈ 1.66666 ≈ 0.6π 3 8. There are many answers to this question. The angles that are co-terminal with an angle of 90◦ can be expressed as x = 90◦ + 360◦ k, kεI where k is any integer. Some examples of the co-terminal angles are x = 90◦ + 360◦ = 450◦ x = 90◦ + 720◦ = 810◦ x = 90◦ − 360◦ = −270◦ x = 90◦ − 720◦ = −630◦ 9. a) There are many answers to this question. The negative angles that are co-terminal with an angle of 120◦ can be expressed as x = 120◦ + 360◦ k, kεI where k is a negative integer. Some examples of the co-terminal angles of 120◦ that are negative angles are: x = 120◦ − 360◦ = −240◦ x = 120◦ − 720◦ = −600◦ b) There are many answers to this question. The angles that are greater than 360◦ and co-terminal with an angle of 120◦ can be expressed as x = 120◦ + 360◦ k, kεI where k is a positive integer. Some examples of the co-terminal angles of 120◦ that are greater than 360◦ are: x = 120◦ + 360◦ = 480◦ x = 120◦ + 720◦ = 840◦ 15 10. 1 πd 2 1 = (π)(240 m) 2 The front outside wheel will complete the most rotations. Ctrack = Ctrack Ctrack 122π = ≈ 203 rotations Cfront wheel 0.6π Ctrack = 120π meters(Inside Distance) 1 πd 2 1 = (π)((244 m) 2 = 122(π)meters Ctrack = Ctrack Ctrack Cfront wheel = πd Cfront wheel = ((π)((0.6m) Cfront wheel = 0.6(π) meters The back inside wheel will complete the least number of rotations Cback wheel = πd Cback wheel = ((π)(1.88m) Cback wheel = 1.88(π) meters Ctrack Cback wheel = 120π 200 ≈ ≈ 67 rotations 1.8π 3 # degrees front tire = 203 rotations × 360◦ # degrees front tire = 73080◦ # degrees back tire = 67 rotations × 360◦ # degrees back tire = 24120◦ # degrees diﬀerence = 73080◦ − 24120◦ # degrees diﬀerence = 48960◦ Deﬁning Trigonometric Functions Review Exercises: 1. In △ABC, with respect to ∠A, the opposite side is 9, the adjacent side is 12, and the hypotenuse is 15. The values of the six trigonometric functions for ∠A are: 16 Table 1.1 Function Ratio Value sin ∠A cos ∠A tan ∠A csc ∠A sec ∠A cot ∠A opp hyp adj hyp opp adj hyp opp hyp adj adj opp 9 15 12 15 9 12 15 9 15 12 12 9 = = = = = = 3 5 4 5 3 4 5 3 5 4 4 3 2. a) In △V ET the hypotenuse is: (h)2 = (s1 )2 + (s2 )2 (h)2 = (8)2 + (15)2 (h)2 = 64 + 225 √ √ h2 = 289 h = 17 b) In △V ET , with respect to T , the opposite side is 15, the adjacent side is 8, and the hypotenuse is 17. The values of the six trigonometric functions for ∠T are: Table 1.2 Function Ratio Value sin ∠T cos ∠T tan ∠T csc ∠T sec ∠T cot ∠T opp hyp adj hyp opp adj hyp opp hyp adj adj opp 15 17 8 17 15 8 17 15 17 8 8 15 3. a) 17 The radius of the circle is (h)2 = (s1 )2 + (s2 )2 (h)2 = (3)2 + (−4)2 (h)2 = 9 + 16 √ √ h2 = 25 ∴ h = 5 With respect to the angle in standard position, θ, the hypotenuse is 5, the opposite is −4, and the adjacent is 3. b) Table 1.3 Function Ratio Value sin θ cos θ tan θ csc θ sec θ cot θ opp hyp adj hyp opp adj hyp opp hyp adj adj opp − 54 3 5 − 34 − 45 5 3 − 43 4. a) (h)2 = (s1 )2 + (s2 )2 The radius of the circle is (h)2 = (−5)2 + (−12)2 (h)2 = 25 + 144 √ (h)2 = 169 ∴ h = 13 With respect to the angle in standard position, θ, the hypotenuse is 13, the opposite is −12, and the adjacent is −5. b) 18 Table 1.4 Function Ratio Value sin θ cos θ tan θ csc θ sec θ cot θ opp hyp adj hyp opp adj hyp opp hyp adj adj opp 12 − 13 5 − 13 −12 −5 13 − 12 − 13 5 −5 −12 = 12 5 = 5 12 5. Table 1.5 Function Value sin θ cos θ tanθ csc θ sec θ cot θ −1 0 undeﬁned −1 undeﬁned 0 6. a) The measure of ∠DAB is 60◦ which is the sum of ∠BAC and ∠DAC. The measure of each angle in △DAB is 60◦ . Therefore the triangle is equiangular. b) The measure of the side BD of △DAB is 1 because it is the third side of △DABwhich is also an equilateral triangle. c) The measure of BC and CD is has a length of one. 1 2 The altitude AC of the equilateral triangle bisects the base BD which d) The ordered pair can be obtained by ﬁrst using the Pythagorean Theorem to determine the measure of AC. 19 (h)2 = (s1 )2 + (s2 )2 (1)2 = (−.5)2 + (s)2 1 = .25 + s2 1 − 0.25 = s2 √ √ 0.75 = s2 √ 3 . 2 ∴ s = 0.8660 which is equivalent to If ∠BAC were represented (√ ) as an angle in standard position, the coordinates on the unit circle would be (cos 30◦ , sin 30◦ ) or 23 , 21 . √ e) If ∠ABC were represented as an angle in standard position, the opposite side would be 3, (and the √ ) adjacent side would be 1. Therefore the coordinates on the unit circle would be (cos 60◦ , sin 60◦ ) or 12 , 23 . 7. (h)2 = (s1 )2 + (s2 )2 (1)2 = (n)2 + (n)2 1 = n2 + n2 1 = 2n2 2n2 1 = 2 √2 √ 1 ± = n2 2 (√ ) √ √ 1 2 2 2 √ = ±√ = ± ±√ 2 2 2 4 √ 2 n= 2 1 ∴ n = ± √ Rationalize the denominator 2 The angle is in the ﬁrst quadrant so the values of (x, y) are positive 8. To determine the values of the six trigonometric functions for 60◦ , the following special triangle may be used. Table 1.6 Function Ratio sin 60◦ cos 60◦ opp hyp adj hyp Value √ 3 2 1 2 20 Table 1.6: (continued) Function Ratio tan 60◦ opp adj hyp opp hyp adj adj opp csc 60◦ sec 60◦ cot 60◦ Value √ 3 1 √2 3 2 1 √1 3 = √ 2 3 3 √ = 3 3 9. An angle in standard position in the ﬁrst quadrant tan θ = oppy adj(x) Since both x and y are positive quantities, then the function will also be positive. An angle in standard position in the third quadrant: tan θ = opp(y) adj(x) Since both x and y are negative quantities, then the function will be positive. 10. An angle of 150◦ drawn in standard position is equivalent to a reference angle of 30◦ drawn in the second quadrant. ( √ ) The coordinates of this angle on the unit circle are (cos 30◦ , sin 30◦ ) which would be − 23 , 21 21 Trigonometric Functions of Any Angle Review Exercises: 1. The reference angle for each of the following angles is: a) 190◦ 190◦ –180◦ = 10◦ b) –60◦ 360◦ –300◦ = 60◦ A negative angle indicates that the angle opens clockwise. c) 1470◦ 1470◦ –4(360◦ ) = 30◦ d) –135◦ 225◦ –180◦ = 45◦ 2. The coordinates for each of the following angles are: a) 300◦ The reference angle is 360◦ –300◦ = 60◦ (4th quadrant) ( √ ) 3 1 (cos 60◦ , sin 60◦ ) = , 2 2 −150◦ The reference angle is 180◦ –150◦ = 30◦ (3rd quadrant) ( √ ) 3 1 (cos 30◦ , sin 30◦ ) = − ,− 2 2 405◦ The reference angle is 405◦ –360◦ = 45◦ (1st quadrant) (√ √ ) 2 2 (cos 45◦ , sin 45◦ ) = , 2 2 b) c) 3. a) sin 210◦ is equivalent to sin 30◦ in the 3rd quadrant. Its value is − 21 . b) tan 270◦ is equivalent to tan 90◦ . Its value is undeﬁned. ◦ c) csc 120◦ is equivalent to csc 60 in the 2nd quadrant. Cosecant is the reciprocal of sine so the value will √ be positive. Its value is √23 = 2 3 3 . 4. a) An angle of 510◦ has a reference angle of 30◦ in the 2nd quadrant. Therefore, the value of sin 510◦ is 12 . b) An angle of 930◦ has a reference angle of 30◦ in the 3rd quadrant. Therefore, the value of cos 930◦ is − 22 √ 3 2 . c) An angle of 405◦ has a reference angle of 45◦ in the 1st quadrant. The value of csc 405◦ is √ 2 1 . 5. a)√An angle of –150◦ has a reference angle of 30◦ in the 3rd quadrant. Therefore the value of cos(−150◦ ) is − 23 . b) An angle of –45◦ has a reference angle of 45◦ in the 4th quadrant. Therefore the value of tan(−45◦ ) is −1. c) An angle of –240◦ has a reference angle of 60◦ in the 2nd quadrant. Therefore the value of sin(−240◦ ) is √ 3 2 . 6. Using the table in the lesson the value of cos 100◦ is approximately −0.1736. 7. Using the table in the lesson, the angle that has a sine value of 0.2 is between 165◦ and 170◦ . 8. The tangent of 50◦ is approximately 1.1918 and this value is very reasonable because tan 45◦ is 1. As the measure of the angle gets larger so does the tangent value of the angle. 9. a) The value of sin 118◦ using the calculator is approximately sin 118◦ ≈ .8829 . b) The value of tan 55◦ using the calculator is approximately tan 55◦ ≈ 1.4281. 23 10. From observing the value displayed in the table, the conjecture that can be made is sin(a) + sin(b) ̸= sin(a + b). 11. This area represents a worksheet for sin(a) and (sina)2 . sin 0◦ = 0 (0)2 = 0 sin 25◦ = 0.4226 √ 2 sin 45◦ = 2 (0.4226)2 = 0.1786 ( √ )2 2 1 = 2 2 sin 80◦ = 0.9848 (0.9848)2 = 0.9698 sin 90◦ = 1 √ (1)2 = 1 ( √ )2 3 3 = 2 4 sin 120◦ = 3 2 sin 235◦ = −0.8192 (−0.8192)2 = 0.6711 sin 310◦ = −0.7660 (−0.7660)2 = 0.5868 This area represents a worksheet for cos(a) and (cos a)2 cos 0◦ = 1 (1)2 = 1 cos 25◦ = 0.9063 √ 2 cos 45◦ = 2 (0.9063)2 = 0.8214 ( √ )2 2 1 = 2 2 cos 80◦ = 0.1736 (0.1736)2 = 0.0301 cos 90◦ = 0 (0)2 = 0 ( )2 1 1 − = 2 4 cos 120◦ = − 1 2 cos 235◦ = −0.5736 (−0.5736)2 = 0.3290 ◦ (0.6428)2 = 0.4132 cos 310 = 0.6428 From the above results the following conjecture can be made: (sin a)2 + (cos a)2 = 1 √ 12. g(x) = 4 + 1 − sin2 x + sin2 x The conjecture that would be made about the value of this function is that it would equal 5. Using the TI-83 to graph the function: 24 In order for this to occur with the above function 1 − sin2 x should be changed to (cos2 x)2 to result in cos2 x + sin2 x which equals one. Relating Trigonometric Functions Review Exercise: Pages 80 – 82 1. a) sec θ = 4 cos θ = 1 sec θ ∴ cos θ = 1 4 b) sin θ = 1 3 csc θ = 1 sin θ csc θ = 1 ∴ csc θ = 3 1 3 2. a) Table 1.7 Angle Sin Csc 10 5 1 0.5 0.1 0 −0.1 −0.5 −1 −5 −10 0.1736 0.0872 0.0175 0.0087 0.0017 0 −0.0017 −0.0087 −0.0175 −0.0872 −0.1736 5.7604 11.4737 57.2987 114.5930 572.9581 undeﬁned −572.9581 −114.5930 −57.2987 −11.4737 −5.7604 b) As the measure of the angle approaches zero degrees, the values of the cosecant increase greatly. c) The value of the sine function has a maximum of one. However, the cosecant function has no maximum value. Its value continues to increase. d) The range of the cosecant function has no values between −1 and +1. However, it does have values from −1 to −∞ and from +1 to +∞. 25 3. Any angles that resulted in a value of zero for the cosine of the angle are excluded from the domain of the secant function. These angles include 90◦ , 270◦ , 450◦ , etc. 4. To answer this question correctly, the following diagram that shows in which quadrant the trigonometric functions are positive, will be used to determine the sign of the given function. S A Sine Cosecant Tangent Cotangent All Cosine Secant T C a) sin 80◦ → The angle is located in the 1st quadrant and its value will be positive. b) cos 200◦ → The angle is located in the 3rd quadrant and its value will be negative. c) cot 325◦ → The angle is located in the 4th quadrant and its value will be negative. d) tan 110◦ → The angle is located in the 2nd quadrant and its value will be negative. 5. cos θ = 6 adj = ; hyp 10 sin θ = opp 8 = ; hyp 10 tan θ = opp 8 4 = = adj 6 3 sin θ 6. In the 3rd quadrant, both the sine function and the cosine function have negative values. tan θ = cos θ θ and cot θ = cos . The result of dividing two negative values is positive. Therefore, in the 3rd quadrant these sin θ quotient identities will have a positive value. 7. All angles in the 1st quadrant have a positive value. sin θ = 0.4 −1 sin (sin θ) = sin−1 (0.4) θ ≈ 23.58◦ Therefore cos 23.58◦ ≈ 0.9165. 8. All angles in the 1st quadrant have a positive value. If cot θ = 2 then tan θ = −1 tan (tan θ) = tan −1 ( ) 1 2 θ ≈ 26.57◦ Therefore csc 26.57◦ ≈ 2.2357 (Note: csc θ = 1 sin θ ). 26 1 2 9. From the above diagram, sin θ = yr ; cos θ = x r and x2 + y 2 = r2 x2 + y 2 = r2 x2 y2 r2 + = y2 r2 r2 2 2 cos θ + sin θ = 1 Dividing through by r2 Replacing the ratios with the correct functions as deﬁned above. The Pythagorean Identity cos2 θ + sin2 θ = 1 can now be used to prove 1 + tan2 θ = sec2 θ. Proof: cos2 θ + sin2 θ = 1 sin2 θ 1 cos2 θ + = cos2 θ cos2 θ cos2 θ 1 + tan2 θ = sec2 θ Dividing through by cos2 θ Using identities for substitutions:. tan θ = sin θ 1 and sec θ = cos θ cos θ 10. It is necessary to indicate the quadrant in which the angle is located in order to determine the correct angle. When using the Pythagorean Identities, the equations are quadratic and a quadratic equation has two possible solutions. If the quadrant is stated in the question, then only one answer is acceptable. Applications of Right Triangle Trigonometry Review Exercises: 1. To solve a triangle means to determine the measurement of all angles and all sides of the given triangle. In △ABC: a ≈ 9.33 b ≈ 5.83 c = 11 ∠A = 58◦ ∠B = 32◦ ∠C = 90◦ ∠A = 180◦ − (32◦ + 90◦ ) ∠A = 58◦ 27 opp hyp b sin 32◦ = 11 b 0.5299 = 11 adj hyp a cos 32◦ = 11 a 0.8480 = 11 sin B = cos B = ( (11)(0.5299) = (11) b 11 ) (11)(0.8480) = (11) 5.83 ≈ b (a) 11 9.33 ≈ a 2. Anna is correct. In order to solve a triangle, the minimum amount of information that must be given is the measure of two angles and one side, or one angle and two sides. 3. (h)2 = (s1 )2 + (s2 )2 (h)2 = (6)2 + (5.03)2 √ √ h2 = 61.3009 4. sin B = 3 5 = 0.6 sin 30◦ = ∴ h ≈ 7.829 ≈ 7.83 This answer conﬁrms those given in example 2. 1 2 = 0.5 Therefore, the measure of ∠B is larger than 30◦ . Using a calculator, sin−1 (sin B) = sin−1 (0.6) ∠B ≈ 36.87◦ ≈ 37◦ 5. 28 opp adj x ◦ tan 53 = 15 x 1.3270 = 15 ( ) x (15)(1.3270) = (15) 15 19.91 feet ≈ x tan ∠A = The length of the ﬂagpole is approximately 19.9 feet. 6. opp adj x tan 76◦ = 30 x 4.0108 = 30 ( ) x (30)(4.0108) = (30) 30 120.32 feet ≈ x tan ∠BAC = The house is approximately 120.3 feet away. 7. 29 opp hyp 200 sin 80◦ = x 200 0.9848 = x opp adj 200 tan 80◦ = x 200 5.6713 = x tan ∠A = sin A = ( (x)(0.9848) = (x) 200 x ) (x)(5.6713) = (x) 0.9848x = 200 200 0.9848x = 0.9848 0.9848 x ≈ 203.09 ≈ 203 miles ( 200 x ) 5.6713x = 200 200 5.6713x = 5.6713 5.6713 x ≈ 35.27 ≈ 35.3 miles The plane has traveled approximately 203 miles. City A and City B are approximately 35.3 miles apart. 8. opp adj x ◦ tan 40 = 50 x 0.8391 = 50 ( ) x (50)(0.8391) = (50) 50 x ≈ 41.96 feet tan ∠C = The lake is approximately 41.96 feet wide. 9. 30 △T AN opp hyp AT sin 50◦ = 3 AT 0.7660 = 3 (3)(0.7660) = (3) 2.29 ≈ 2.3 ≈ AT adj hyp NT cos 50◦ = 3 NT 0.6428 = 3 cos ∠N = sin N = ( AT 3 ) (3)(0.6428) = (3) ( NT 3 ) 1.93 ≈ 1.9 ≈ N T △P AT PT = NP − NT (h)2 = (s1 )2 + (s2 )2 P T = 9.0 − 1.9 P T = 7.1 (h)2 (7.1)2 + (2.3)2 √ √ h2 = 55.7 ∴ h ≈ 7.46 AT = 2.3 The length of side x is approximately 7.46 31 32 Chapter 2 TE Circular Functions - Solution Key 2.1 Circular Functions Radian Measure Review Exercises 1. a) The circle that is missing appears to be one-third of the circle. Therefore the measure of the angle could be estimated to be 120◦ . b) 120◦ · π 180◦ = 120◦ π 180◦ = 2π 3 radians c) The part of the cheese that remains has a measure of 360◦ − 120◦ = 240◦ . 240◦ · π 240◦ π 4π = = radians 180◦ 180◦ 3 2. Table 2.1 Angle in Degrees Radian Measure 240◦ 270◦ 315◦ −210◦ 120◦ 15◦ −450◦ 72◦ 720◦ 330◦ π 240 π 4π 240◦ · 180 ◦ = 180◦ = 3 radians ◦ π 270 π 3π 270◦ · 180 ◦ = 180◦ = 2 radians π 315◦ π ◦ 315 · 180◦ = 180◦ = 7π 4 radians π −210◦ π 7π −210◦ · 180 = = ◦ 180◦ 6 radians ◦ 120 π 2π π 120◦ · 180◦ = 180◦ = 3 radians π 15◦ π π 15◦ · 180 ◦ = 180◦ = 12 radians −450◦ π π ◦ −450 · 180◦ = 180◦ = − 5π 2 radians 72◦ π 2π π 72◦ · 180 ◦ = 180◦ = 5 radians 720◦ π π 720◦ · 180 ◦ = 180◦ = 4π radians π 330◦ π 11π 330◦ · 180 ◦ = 180◦ = 6 radians ◦ 33 3. Table 2.2 Angle in Radians Degree Measure π 2 11π 5 2π 3 ◦ π2 · 180 = 180 2 = 90 π ◦ ◦ 1980 11π 180 = 396◦ 5 · π = 5 ◦ ◦ 2π ◦ 3 · 180 = 360 3 = 120 π◦ 5 π · 180 = 900◦ π ◦ 7 π 180 1260◦ = 630◦ 2 · π = 2 ◦ ◦ 3 π 180 540 ◦ 10 · π = 10 = 54 ◦ ◦ 5 π 180 900 ◦ 12 · π = 12 = 75 ◦ ◦ − 136π 180 = 2340 = −390◦ 6 π◦ 8 π · 180 = 1440◦ π ◦ 4 π 180 720◦ ◦ 15 π = 15 = 48 ◦ 5π 7π 2 3π 10 5π 12 − 13π 6 8π 4π 15 4. 5. 34 ◦ a) 6π 7 rad = b)1rad = 6(180◦ ) 7 180◦ π ≈ 154.3◦ ≈ 57.3◦ c) 3rad = 57.3◦ . 3 ≈ 171.9◦ d) 20π 11 = 20(180◦ ) 11 ≈ 327.3◦ 6. a) sin 210◦ = − 12 b) Gina calculated sin 210 wit her calculator in radian mode. 7. Table 2.3 Angle(x) Sin(x) Cos(x) √ 5π ◦ ◦ 4 (225 → 45 ) 11π ◦ ◦ 6 (330 → 30 ) 2π ◦ ◦ 3 (120 → 60 ) π ◦ 2 (90 ) 7π ◦ ◦ 2 (630 → 270 ) Tan(x) √ − 22 1 − √2 3 2 2 − √ 2 3 2 − 12 1 −1 0 0 1√ − 33 √ − 3 undeﬁned undeﬁned Applications of Radian Measure Review Exercises 1. a) 360◦ 24 b) π 12 = 15◦ 15◦ · π 180◦ = 15π 180 π 12 rad = ≈ 0.3rad ◦ c) 15 2. a) 360◦ 12 = 30◦ 30◦ ( π 180◦ ) = 30π 180 = π 6 rad b) π (0.5m) ≈ 0.262m 6 0.262m · 100cm/m ≈ 26cm 3. a) 360◦ 32 = 45◦ 4 45◦ 4 ( π 180◦ ) = 45π 720 = π 16 rad π b) The distance between two consecutive dots on the circle is 16 rad. Since the chord spans 13 dots, the 13π measure of the central angle is 16 rad The length of the chord is: 35 c = 2r sin θ 2 ( ) 13π 1 16 2 13π c = 2(1.20m) sin 32 c ≈ 2.297 ≈ 2.3m c ≈ 2.3m (100cm/m) ≈ cm c = 2(1.20m) sin 4. a) 360◦ 12 = 30◦ 30◦ ( π 180◦ ) = 30π 180 = π 6 rad The area of each designated is equal to the area of the outer sector – the area of the inner sector. 1 1 A(outer) r2 θ − A(inner) r2 θ 2 2 (π ) (π ) 1 1 A(outer) (110)2 − A(inner) (55)2 2 6 2 6 ≈ 3167.77 − 791.94 ≈ 2375.83 ≈ 2376 ft2 The approximate area of each section is 2376 ft2 . The students from Archimedes High school have four allotted sections: 4(2376 ft2 ) = 9504 ft2 b) There are three sections allotted for general admission: 3(2376 ft2 ) = 7128 ft2 c) The press and the oﬀicials have one allotted section: 2376 ft2 5. Diameter of the gold circle: Radius of the gold circle: 11 2 Diameter of the red circle: Radius of the red circle: 22 2 1 3 (33) = 11 inches = 5.5 inches 2 3 (33) = 22 inches = 11 inches Step One: A(total red) πr2 − A(gold) πr2 A(total red) π(11)2 − A(gold) π(5.5)2 ≈ 285.1 inches2 36 Step Two: 1 1 A(red sector) r2 θ − A(gold sector) r2 θ 2 2 ( ) (π) 1 1 2 π A(red sector) (11) − A(gold sector) (5.5)2 2 4 2 4 ≈ 35.6 inches2 Step Three: 285.1 − 35.6 ≈ 249.5 inches2 Circular Functions of Real Numbers Review Exercises: 1. Using similar triangles: 37 x 1 = 1 A A= Ax = 1 cos θ = x 1 = sec θ cos θ 1 cos θ ∴ A = sec θ Ax 1 = x x ∴A= 1 x 2. (h)2 = (s1 )2 + (s2 )2 (sec θ)2 = (1)2 + (tan θ)2 sec2 θ = 1 + tan2 θ 38 3. 39 4. 5. The tan(x) and sec(x) are two trigonometric functions that increase as x increases from 0 to 6. As x increases from 3π 2 to 2π, cot(x) gets inﬁnitely smaller. Linear and Angular Velocity Review Exercises 1. a) s t 43.98 v= 9 v ≈ 4.89 cm/sec c = 2πr v= c = 2π(7 cm.) c ≈ 43.98 cm b) w = θ t (where θ is one rotation (2π) and t is the time to complete 1 rotation) 2π 9 w ≈ 0.698 w ≈ 0.70 rad/sec w= 40 π 2. 2. a) s t 43.98 v= 3.5 v ≈ 12.57 cm/sec v= b) w = θ t (where θ is one rotation (2π) and t is the time to complete 1 rotation) 2π 3.5 w ≈ 1.795 w= w ≈ 1.80 rad/sec 3. a) w = θ t (where θ is one rotation (2π) and t is the time to complete 1 rotation) 2π 12 w ≈ 524 w= w ≈ 0.524 rad/sec b) w = θ t Velocity for Lois: Velocity for Doris: v = rw v = (3m)(0.524) v ≈ 1.57 m/sec v = rw v = (10m)(0.524) v ≈ 5.24 m/sec (where θ is one rotation (2π) and t is the time to complete 1 rotation) 2π 12 w ≈ 524 w ≈ 0.524 rad/sec w= 4. a) s t s t= v 2.7 × 104 t= 3 × 108 t ≈ .9 × 10−4 ≈ 9.0 × 10−5 seconds v= 41 b) w = θ t (where θ is one rotation (2π) and t is the time to complete 1 rotation) 2π 9.0 × 10−5 w ≈ 69813.17 rad/sec ≈ 69813 rad/sec w= c) # rotations # rotations # rotations v where v is the speed of the protons and c is the circumference of the LHC. c 3 × 108 = 2700 ≈ 11, 111 rotations in 1 second = Graphing Sine and Cosine Functions Review Exercises 1. The graph of y = sec(x) The period is 2π and the frequency is 1. The graph of y = cot(x) The period is π and the frequency is 2. 2. Table 2.4 Function Minimum Value Maximum Value a) y = cos x b) y = 2 sin x c) y = − sin x d) y = tan x −1 −2 −1 −∞ 1 2 1 +∞ 42 3. For the equation 4 sin(x) = sin(x) over the interval 0 ≤ x ≤ 2π there are 3 real solutions. 4. Table 2.5 Function y y y y y y = cos(2x) = 3 sin x = 2 sin(πx) = 2 cos(3x) ( ) = 21 cos( 12 x) = 3 sin 21 x Period Amplitude Frequency π 2π 2π 3 2π 3 1 3 2 2 2 1 3 3 4π 4π 1 2 3 1 2 1 2 a) y = cos(2x) The period is π. This is the interval required to graph one complete cosine curve. The amplitude is the distance from the sinusoidal axis to the maximum point of the curve. The amplitude of y = cos(2x) is 1. The frequency is the number of complete curves that are graphed over the interval of 2π. The frequency for y = cos(2x) is 2. b) y = 3 sin x The period is 2π. This is the interval required to graph one complete sine curve. The amplitude is the distance from the sinusoidal axis to the maximum point of the curve. The amplitude of y = 3 sin(x) is 3. The frequency is the number of complete curves that are graphed over the interval of 2π. The frequency for y = 3 sin(x) is 1. c) y = 2 sin(πx) 43 The period is 2π 3 . This is the interval required to graph one complete sine curve. The amplitude is the distance from the sinusoidal axis to the maximum point of the curve. The amplitude of y = 2 sin(πx) is 2. The frequency is the number of complete curves that are graphed over the interval of 2π. The frequency for y = 2 sin(πx) is 3. d) y = 2 cos(3x) The period is 2π 3 . This is the interval required to graph one complete cosine curve The amplitude is the distance from the sinusoidal axis to the maximum point of the curve. The amplitude of y = 2 cos(3x) is 2. The frequency is the number of complete curves that are graphed over the interval of 2π. The frequency for y = 2 cos(3x) is 3. ( ) e) y = 12 cos 12 x Graph over 2π The period is 4π. This is the interval required to graph one complete cosine curve The amplitude ( is ) the distance from the sinusoidal axis to the maximum point of the curve. The amplitude of y = 12 cos 12 x is 12 . 44 Graph over 4π The frequency ( ) is the number of complete curves that are graphed over the interval of 2π. The frequency for y = 12 cos 12 x is 12 . ( ) f) y = 3 sin 12 x Graph over 2π The period is 4π. This is the interval required to graph one complete sine curve. The amplitude ( is) the distance from the sinusoidal axis to the maximum point of the curve. The amplitude of y = 3 sin 12 x is 3. The frequency ( ) is the number of complete curves that are graphed over the interval of 2π. The frequency for y = 3 sin 12 x is 12 . 5. Table 2.6 Period Amplitude Frequency Equation π 4π 3 2 2 2 y y y y π 2 π 3 1 2 4 6 1 2 6. a) y = 3 sin(2x) 45 =3 =2 =2 = 12 cos(2x) ( ) sin 12 x cos(4x) sin(6x) b) y = 2.5 cos(πx) c) y = 4 sin (1 ) 2x Translating Sine and Cosine Functions Review Exercises 1. B the minimum value is 0. ( ) A. y = sin x + π2 2. E the maximum value is 3. B. y = 1 + sin(x) 3. D the minimum value is − 2. C. y = cos(x − π) ( ) 4. C the y-intercept is − 1. D. y = −1 + sin x − 3π 2 5. A the same graph as y = cosx E. y = 2 + cos x ( ) 6. y = −2 + sin(x + π) and y = −2 + cos x + π2 ( ) 7. y = 2 + sin x − π2 Graph C ( ) 8. y = −1 + cos x + 3π Graph D 2 ( ) 9. y = 2 + cos x − π2 Graph A 10. y = −1 + sin(x − π) Graph B ( ) 11. The graph of y = 1 + sin x − π4 46 General Sinusoidal Functions Review Exercises The following general form of a sinusoidal function will be used to answer 1 – 5. y = C + A sin(B(x − D)) where: C represents the Vertical Translation(V.T.) A represents the Vertical Stretch (amplitude) (V.S.) B represents the Horizontal Stretch (H.S.) D represents the Horizontal Translation (H.T.) 1. y = 2 + 3 sin(2(x − 1)) The graph of this sinusoidal curve is the graph of y = sin x that has been vertically translated upward 2 units and horizontally translated I unit to the right. The amplitude of the curve is 3 and the period is 12 (2π) or π. The frequency is 2. The graph will have a maximum value of 5 and a minimum value of −1. 2. y = −1 + sin(π(x + π3 )) The graph of this sinusoidal curve is the graph of y = sin x that has been vertically translated downward 1 unit and horizontally translated π3 units to the left. The amplitude of the curve is 1 and the period is 2. The frequency is π. The graph will have a maximum value of 0 and a minimum value of −2. 3. y = cos(40x − 120) + 5 The graph of this sinusoidal curve is the graph of y = cos x that has been vertically translated upward 5 units and horizontally translated 30 radians to the right. The amplitude of the curve π is 1 and the period is 20 . The frequency is 40. The graph will have a maximum value of 5 and a minimum value of 4. ( ( )) 4. y = − cos 21 x + 5π The graph of this sinusoidal curve is the graph of y = cos x that has not been 4 vertically translated but has been horizontally translated 5π 4 radians to the left. The negative sign in front of the function indicates that the graph has been reﬂected across the x−axis. The amplitude of the curve is 1 and the period is 4π. The frequency is 21 . The graph will have a maximum value of 1 and a minimum value of −1. 5. y = 3 + 2 cos(−x) The graph of this sinusoidal curve is the graph of y = cos x that has been vertically translated upward 3 units. There is no horizontal translation. However, the negative sign in front of the x indicates that the graph has been reﬂected across the y−axis. This reﬂection is not visible in the graph 47 since the graph is symmetric with the y−axis. The amplitude of the curve is 2 and the period is 2π. The frequency is 1. The graph will have a maximum value of 5 and a minimum value of 1. All of the above answers can be conﬁrmed by using the TI-83 to graph each function. 6. For this graph, the transformations of y = cos(x) are: V R → N o; V S → 2; V T → 3 ( ) π 1 1 π HS → = ; HT → 2 2π 4 6 ( ( )) The equation that models the graph is y = 3 + 2 cos 4 x − π6 7. For this graph, the transformations of y = sin(x) are: V R → N o; V S → 1; V T → 2 2π 3π HS → = 1; HT → − 2π 2 ( The equation that models the graph is y = 2 + sin x + 3π 2 ) 8. For this graph, the transformations of y = cos(x) are: V R → N o; V S → 20; V T → 10 60◦ 1 HS → = ; HT → 30◦ ◦ 360 6 The equation that models the graph is y = 10 + 20 cos(6(x − 30◦ )) 9. For this graph, the transformations of y = sin(x) are: V R → N o; V S → HS → 3 ;V T → 3 4 4π = 2; HT → −π 2π The equation that models the graph is y = 3 + 3 4 sin (1 2 (x ) + π) 10. For this graph, the transformations of y = cos(x) are: V R → N o; V S → 7; V T → 3 12π π HS → = 3; HT → 4π 4 The equation that models the graph is y = 3 + 7 cos (1 ( 3 48 x− π 4 )) Chapter 3 TE Trigonometric Identities Solution Key 3.1 Trigonometric Identities Fundamental Identities Review Exercises: 1. If the tangent of an angle has a negative value, then the angle must be found in either the 2nd or 4th quadrant. If cos θ > 0, then the angle must be located in the 4th quadrant since the cosine function is positive in this quadrant. Given θ = − 32 and this angle is located in the 4th quadrant, the negative value is 2. To determine the value of sin θ, the length of the hypotenuse must be found by using Pythagorean Theorem. (h)2 = (s1 )2 + (s2 )2 (h)2 = (2)2 + (−3)2 √ √ h2 = 13 √ h = 13 49 sin θ = opp hyp −2 −2 sin θ = √ → sin θ = √ 13 13 √ 2 13 sinθ = − 13 (√ 13 √ 13 ) 2. If csc θ = −4 and sin θ = csc1 θ then sin θ = − 14 . The sine function is negative in the 3rd and 4th quadrants. However, if tan θ > 0, then the angle must be in the 3rd quadrant since the value of the tangent function is positive in this quadrant. The length of the adjacent side must be found by using Pythagorean Theorem. (h)2 = (s1 )2 + (s2 )2 (4)2 = (−1)2 + (s2 )2 √ √ sin θ = opp hyp cos θ = 16 = 1 + (s2 )2 √ 15 = s2 15 = s adj hyp √ 1 sin θ = − 4 15 cos θ = − 4 csc θ = −4 4 sec θ = − √ → 15 √ 15 cot θ = 1 √ 4 15 sec θ = − 15 tan θ = opp adj 1 1 tan θ = √ → tan θ = √ 15 15 √ 15 tan θ = 15 ( √ √ ) 4 15 √ sec θ = − √ 15 15 (√ 15 √ 15 ) 3. If sin θ = 13 , then the angles are located in the 1st and 2nd quadrants since the sine function is positive in these quadrants. There are also two values for the cosine function in these quadrants. The length of the adjacent side must be found by using Pythagorean Theorem. 50 (h)2 = (s1 )2 + (s2 )2 (3)2 = (1)2 + (s2 )2 √ 9 = 1 + (s2 )2 √ √ 8 = s2 4·2=s √ 2 2=s adj hyp √ 2 2 cos θ = 3 adj hyp √ 2 2 cos θ = − 3 cos θ = cos θ = 4. If cos θ = − 25 and the angle is located in the 2nd quadrant, the length of the opposite side must be determined in order to determine the values of the remaining trigonometric functions. (h)2 = (s1 )2 + (s2 )2 (5)2 = (−2)2 + (s2 )2 √ √ 25 = 4 + (s2 )2 √ 21 = s2 21 = s 51 opp hyp √ 21 sin θ = 5 sin θ = sec θ = hyp adj 5 sec θ = − 2 tan θ = opp adj √ csc θ = 21 tan θ = − 2 cot θ = hyp opp 5 5 csc θ = √ = √ 21 21 adj opp 2 2 cot θ = − √ = − √ 21 21 (√ 21 √ 21 ) (√ 21 √ 21 ) = √ 5 21 21 √ 2 21 =− 21 5. If (3, −4) is on the terminal side of the angle in standard position, the angle is located in the 4th quadrant. The Pythagorean Theorem can be used to determine the length of the hypotenuse. (h)2 = (s1 )2 + (s2 )2 (h)2 = (3)2 + (−4)2 (h)2 = 9 + 16 √ √ h2 = 25 h=5 opp hyp 4 sin θ = − 5 sin θ = adj hyp 3 cos θ = 5 cos θ = opp adj 4 tan θ = − 3 hyp opp 5 csc θ = − 4 tan θ = csc θ = hyp adj 5 sec θ = 3 sec θ = adj opp 3 cot θ = − 4 cot θ = 6. If (2, 6) is on the terminal side of the angle in standard position, the angle is located in the 1st quadrant. The values of the trigonometric functions will all be positive. The Pythagorean Theorem can be used to determine the length of the hypotenuse. 52 (h)2 = (s1 )2 + (s2 )2 (h)2 = (2)2 + (6)2 (h)2 = 40 √ √ h2 = 40 √ √ h2 = 4 · 10 √ h = 2 10 sin θ = opp hyp cos θ = 6 6 sin θ = √ = √ 2 10 2 10 tan θ = tan θ = opp adj 6 =3 2 (√ 10 √ 10 ) √ 3 10 = 10 adj hyp 2 2 cos θ = √ = √ 2 10 2 10 hyp csc θ = opp √ √ 2 10 10 csc θ = = 6 3 hyp sec θ = adj √ √ 2 10 sec θ = = 10 2 7. a) 53 (√ 10 √ 10 ) √ = 10 10 cot θ = cot θ = adj opp 2 1 = 6 3 opp hyp 12 sin θ = 13 adj hyp 5 cos θ = 13 sin θ = cos θ = ( sin2 θ + cos2 θ = 1 )2 ( )2 12 5 + =1 13 13 144 25 + =1 169 169 169 =1 169 1=1 b. sin θ = opp hyp sin θ = 1 2 adj hyp √ 3 cos θ = 2 cos θ = sin2 θ + cos2 θ = 1 ( )2 ( √ )2 1 3 + =1 2 2 1 3 + =1 4 4 4 =1 4 1=1 8. a) To factor sin2 θ −cos2 θ, use the diﬀerence of squares. If this does not appear to be an obvious approach, let x2 = sin2 θ and let y 2 = cos2 θ and factor x2 − y 2 . 54 sin2 θ − cos2 θ x2 − y 2 (sin θ + cos θ)(sin θ − cos θ) (x + y)(x − y) → √ x2 = sin2 θ → x = sin θ √ √ → y 2 = cos2 θ → y = cos θ √ (sin θ + cos θ)(sin θ − cos θ) b) sin2 θ + 6 sin θ + 8 (sin θ + 4)(sin θ + 2) 9. sin4 θ−cos4 θ sin2 θ−cos2 θ To simplify this expression, the ﬁrst step is to factor the expression. (sin2 θ + cos2 θ)(sin2 θ − cos2 θ) (sin2 θ − cos2 θ) (sin2 θ − cos2 θ) = 1 → substitute(sin2 θ + cos2 θ = 1) (sin2 θ − cos2 θ) 10. tan2 θ + 1 = sec2 θ To prove this identity, use the quotient identity tan θ = identity sec θ = cos1 θ sin2 θ 1 +1= 2 cos θ cos2 θ ( 2 ) 2 cos θ 1 sin θ +1 = cos2 θ cos2 θ cos2 θ ( 2 ) sin2 θ cos θ 1 + = 2 2 cos θ cos θ cos2 θ sin θ cos θ and the reciprocal sin2 θ + cos2 θ 1 = 2 cos θ cos2 θ 1 1 = → substitute(sin2 θ + cos2 θ = 1) cos2 θ cos2 θ Verifying Identities Review Exercises: To verify a trigonometric identity, it is often easier to work with only one side of the given equation. The goal will then be to have the left side read the same as the right side. Working with only one side of the equation means that the solution is always visible and there is no confusion as to what is being sought. This method will not work 100% of the time but it will work a lot of the time. If one side is kept constant, then all manipulations can be done to achieve the same constant on the working side. 1. Verify sin x tan x + cos x = sec x 55 sin x tan x + cos x = sec x ( ) sin x sin x sin x + + cos x = sec x → tan x = cos x cos x sin2 x + cos x = sec x cos2 x sin2 x ( cos x ) + cos x = sec x → common deno min ator cos2 x cos x sin2 x cos2 x + = sec x cos x cos x sin2 x + cos2 x = sec x → sin2 x + cos2 x = 1 cos x 1 1 = sec x → = sec x cos x cos x sec x = sec x 2. Verify cos x − cos x sin2 x = cos3 x cos x − cos x sin2 x = cos3 x → remove the common factor cos x cos x(1 − sin2 x) = cos3 x → sin2 x + cos2 x = 1 → cos2 x = 1 − sin2 x cos x(cos2 x) = cos3 x cos3 x = cos3 x 3. Verify sin x 1+cos x + 1+cos x sin x = 2 csc x sin x 1 + cos x + = 2 csc x → (common deno min ator) 1 + cos x sin x ( ) ( ) sin x sin x 1 + cos x 1 + cos x + = 2 csc x → expand sin x 1 + cos x 1 + cos x sin x sin2 x 1 + 2 cos x + cos2 x + = 2 csc x → rearrange (sin x)(1 + cos x) (sin x)(1 + cos x) sin2 x + cos2 x + 1 + 2 cos x = 2 csc x → (sin2 x + cos2 x = 1) (sin x)(1 + cos x) 1 + 1 + 2 cos x = 2 csc x → simplify (sin x)(1 + cos x) 2 + 2 cos x = 2 csc x → (remove common factor) (sin x)(1 + cos x) 2(1 + cos x) = 2 csc x → simplify (sin x)(1 + cos x) 2 = 2 csc x (sin x) 1 1 2 = 2 csc x → csc x = (sin x) sin x 2 csc x = 2 csc x 56 4. Verify sin x 1+cos x = 1−cos x sin x 1 − cos x sin x = → cross multiply 1 + cos x sin x (sin x)(sin x) = (1 − cos x)(1 + cos x) sin2 x = 1 − cos2 x → sin2 x + cos2 x = 1 → sin2 x = 1 − cos2 sin2 x = sin2 x 5. Verify 1 1+cos a + 1 1−cos a = 2 + 2 cot2 a 1 1 + 1 + cos a 1 − cos a ( ) ( ) 1 − cos a 1 1 + cos a 1 + 1 − cos a 1 + cos a 1 + cos a 1 − cos a 1 − cos a 1 + cos a + 2 1 − cos a 1 − cos2 a 1 − cos a + 1 + cosa 1 − cos2 a = 2 + 2 cot2 a → (common deno min ator) = 2 + 2 cot2 a → multiply = 2 + 2cot2 a → simplify = 2 + 2cot2 a → simplify → sin2 a + cos2 a = 1 → sin2 a = 1 − cos2 a 2 sin2 2 sin2 2 sin2 2 sin2 2 sin2 2 sin2 2 sin2 2 sin2 = 2 + 2 cot2 a → (remove common factor) cos a = 2(1 + cot2 a) → cot a = sin a ( ) cos2 a =2 1+ → (common deno min ator) sin2 (( 2 ) ) sin a cos2 a =2 1 + → multiply sin2 a sin2 a ( 2 ) sin a cos2 a + =2 → simplify sin2 a sin2 a ( 2 ) sin a + cos2 a =2 → sin2 a + cos2 a = 1 sin2 a ) ( 1 =2 sin2 a 2 = sin2 a 6. Verify cos4 b − sin4 b = 1 − 2 sin2 b cos4 b − sin4 b = 1 − 2 sin2 b → factor (cos2 b − sin2 b)(cos2 b + sin2 b) = 1 − 2 sin2 b → sin2 a + cos2 a = 1 (cos2 b − sin2 b) = 1 − 2 sin2 b → sin2 b + cos2 b = 1 → cos2 b = 1 − sin2 b (1 − sin2 b − sin2 b) = 1 − 2 sin2 b → simplify 1 − 2 sin2 b = 1 − 2 sin2 b 57 7. Verify sin y+cos y sin y − cos y−siny cosy = sec y csc y sin y + cos y cos y − sin y − = sec y csc y → common denominator sin y cos y ( ) ( ) cos y sin y + cos y sin y cos y − sin y − = sec y csc y → nultiply cosy sin y sin y cos y cos y sin y + cos2 y cos y sin y + sin2 y − cos y sin y cos y sin y cos y sin y − cos y sin y + sin2 y + cos2 y cos y sin y 1 cos y sin y 1 1 · cos y sin y sec y csc y 8. Verify (sec x − tan x2 )2 = = sec y csc y → rearrange = sec y csc y → simplify → sin2 y + cos2 y = 1 1 as factors cos y sin y 1 1 = sec csc y → = sec y and = csc y cos y sin y = sec y csc y = sec y csc y → express 1−sin x 1+sin x 1 − sec x → expand 1 + sin x 1 − sec x 1 sin x sec2 x − 2 sec x tan x + tan2 = → sec x = and tan x = 1 + sin x cos x cos x ( )( ) 1 1 sin x sin2 x 1 − sec x − 2 + = → simplify 2 cos x cos x cos x cos2 x 1 + sin x (sec x − tan x)2 = 1 sin x sin2 x 1 − sin x −2 2 + = → simplify 2 cos x cos x cos2 x 1 + sin x 1 − 2 sin x + sin2 x 1 − sin x = → factor cos2 x 1 + sin x → sin2 x + cos2 x = 1 − sin2 x (1 − sin x)(1 − sin x) 1 − sin x = → factor 2 1 + sin x 1 − sin x 1 − sin x (1 − sin x)(1 − sin x) = → simplify (1 − sin x)(1 + sin x) 1 + sin x 1 − sin x 1 − sin x = 1 + sin x 1 + sin x 9. To show that 2 sin x cos x = sin 2x for unit circle to simplify the expression. 5π 6 substitute this value in for x and then use the values from the 58 2 sin x cos x = sin 2x ( ) 5π 5π 5π 5π 1 5π 2 sin cos = sin 2 → = ; cos 6 6 6 6 2 6 √ ( ) ( √ ) 1 3 5π 5π 3 2 · − = sin → sin =− 2 2 6 3 2 ( √ ) √ 3 3 =− → simplify 2 − 4 2 √ √ 3 3 − =− 2 2 √ =− 3 ;2 2 ( 5π 6 ) = 5π 3 10. Verify sec x cot x = csc x 1 cos x sec x cot x = csc x → sec x = ; cot x = cos x sin x 1 ( cos x ) = csc x → simplify cos x sin x 1 1 = csc x → csc x = sin x sin x csc x = csc x Sum and Diﬀerence Identities for Cosine Review Exercises: Pages 246 – 250 1. To calculate the exact value of cos 5π 12 , the angle must be expressed in the form of the sum of two special angles. Once this is done, then the exact value can be determined by using the values for these angles. The unit circle can be used to determine these values. (Hint: It may be easier to convert the measure of the angle to degrees) ◦ = 5(180) = 75◦ Two special angles that add to 12 ( π These ( 750 )are π450 and 300. ) π values can now be converted π and 30 back to radians or the degrees may be used. 45◦ 180 = ◦ 4 180◦ = 6 5π 12 59 5π ( π π ) = + 12 4 6 cos(a + b) = cos a cos b − sin a sin b (π π) ( π)( π) ( π)( π) cos + = cos cos − sin sin 4 6 4 ( 4 6 )6 ( )( ) ( π π ) ( 1 ) √3 1 1 cos + = √ − √ 4 6 2 2 2 2 √ (π π) 3 1 + = √ − √ cos 4 6 2 2 2 2 √ (π π) 3 1 + cos = √ − √ 4 6 2 2 2 2 ( π π ) √3 − 1 √ cos + = → rationalize denominator 4 6 2 2 (√ ) (√ ) (π π) 2 3−1 √ √ cos + = 4 6 2 2 2 √ √ ( π π ) √6 − √2 6− 2 √ cos + = = → simplify 4 6 4 2 4 2. To begin this question, sketch the two angles in standard position and use the Pythagorean Theorem to calculate the length of the adjacent side. (h)2 = (s1 )2 + (s2 )2 (h)2 = (s1 )2 + (s2 )2 (13)2 = (12)2 + (s2 )2 (5)2 = (3)2 + (s2 )2 169 = 144 + (s2 )2 √ √ 25 = s2 25 = 9 + (s2 )2 √ √ 16 = s2 5=s 4=s Now, the value of cos(y − z) can be determined. 60 cos(y − z) = cos y cos z + sin y sin z ( )( ) ( )( ) 5 4 12 3 cos(y − z) = − + 13 5 13 5 20 36 cos(y − z) = − + 65 65 16 cos(y − z) = 65 3. There is more than one combination that could be used to determine the exact value of 345◦ . Two possible combinations are: 345◦ = (300◦ + 45◦ )345◦ = (120◦ + 225◦ ). Both of these will result in the same solution. cos(a + b) = cos a cos b − sin a sin b cos(120◦ + 225◦ ) = cos 120◦ cos 225◦ − sin 120◦ sin 225◦ ( ) (√ ) ( ) )( 1 3 1 1 ◦ ◦ cos(120 + 225 ) = − − −√ −√ 2 2 2 2 √ 1 3 cos(120◦ + 225◦ ) = √ + √ 2 2 2 2 √ 1+ 3 ◦ ◦ √ cos(120 + 225 ) = → rationalize denominator 2 2 (√ ) ( √ ) 2 1+ 3 ◦ ◦ √ cos(120 + 225 ) = √ 2 2 2 √ √ 2+ 6 ◦ ◦ √ cos(120 + 225 ) = → simplify 2 4 √ √ 2+ 6 √ cos(120◦ + 225◦ ) = 2 4 4. cos 80 cos 20 + sin 80 sin 20 is the result of cos(a − b) = cos a cos b + sin a sin b. Therefore the angle is cos(80 − 20) = cos 60. The value of cos 60 is 21 . 3π 4π 5. The exact value of cos 7π 12 determined by calculating the sum of 12 and 12 . These are two of the special 3π π 4π π angles. The angle 12 = 4 and the angle 12 = 3 . Therefore, the exact value can be determined by using the cosine identity for the sum of two angles: cos(a + b) = cos a cos b − sin a sin b 61 cos(a + b) = cos a cos b − sin a sin b (π π) ( π)( π) ( π)( π) cos + = cos cos − sin sin 4 3 4 3 (4 ) 3 ( π π ) ( 1 ) ( 1 ) ( 1 ) √3 + = √ → multiply cos − √ 4 3 2 2 2 2 √ (π π) 1 3 cos + = √ − √ → rationalize deniminator 4 3 2 2 2 2 (√ ) ( ) (√ ) ( √ ) (π π) 2 1 2 3 √ √ = √ cos + − √ → simplify 4 3 2 2 2 2 2 2 √ √ (π π) 2 6 + = √ − √ cos 4 3 2 4 2 4 ( π π ) √2 − √6 cos + = 4 3 4 6. Verify cos(m−n) sin m cos n = cot m + tan n To verify this identity cos(m − n) must be expanded using the cosine identity for the diﬀerence of two angles. In addition, cot m and tan n must be expressed in terms of sine and cosine. The next step will be to work with the right side of the equation so that it reads the same as the left side. cos(m − n) sin m cos n cos m cos n + sin m sin n sin m cos n cos m cos n + sin m sin n sin m cos n cos m cos n + sin m sin n sin m cos n cos m cos n + sin m sin n sin m cos n = cot m + tan n cos m sin n + → common denominator (RS) sin m cos n ( cos n ) ( cos m ) ( sin m ) ( sin n ) = + → multiply cos n sin m sin m cos n cos m cos n sin m sin n = + → simplify sin m cos n sin m cos n cos m cos n + sin m sin n = sin m cos n = 7. Prove cos(π + θ) = − cos θ To prove the above expression is simply a matter of using the cosine identity for the sum of two angles. 62 cos(a + b) = cos a cos b − sin a sin b cos(π + θ) = cos π cos θ − sin π sin θ cos(π + θ) = (−1) cos θ − (0) sin θ cos(π + θ) = − cos θ 8. Verify cos(c+d) cos(c−d) = 1−tan c tan d 1+tan c tan d To verify this identity, the cosine identity for both the sum and the diﬀerence of angles must be used. As well, the quotient identity for tangent must be applied. 1 − tan c tan d cos(c + d) = cos(c − d) 1 + tan c tan d cos c cos d − sin c sin d 1 − tan c tan d = → ÷(LS) by cos c cos d cos c cos d + sin c sin d 1 + tan c tan d cos c cos d−sin c sin d cos c cos d+cos c cos d cos c cos d−sin c sin d cos c cos d+cos c cos d = 1 − tan c tan d sin c sin d → = tan c = tan d 1 + tan c tan d cos c cos d 1 − tan c tan d 1 − tan c tan d = 1 + tan c tan d 1 + tan c tan d 9. To show that cos(a + b) · cos(a − b) = cos2 a − sin2 b, the cosine identity for both the sum and the diﬀerence of angles must be used. Then the Pythagorean identity sin2 θ + cos2 θ = 1, or a form of this identity, will be applied. cos(a + b) · cos(a − b) = cos2 a − sin2 b cos(a + b) · cos(a − b) = cos2 a − sin2 b (cos a cos b − sin a sin b)(cos a cos b + sin a sin b) = cos2 a − sin2 b → multiply cos2 a cos2 b − sin a sin b cos a cos b + sin a sin b cos a cos b − sin2 a sin2 b → simplify cos2 a cos2 b − sin2 a sin2 b = cos2 a − sin2 b → cos2 b = 1 − sin2 b → sin2 a = 1 − cos2 a cos2 a(1 − sin2 b) − (1 − cos2 a) sin2 b = cos2 a − sin2 b → expand cos2 a − cos2 a sin2 b − sin2 b + cos2 a sin2 b = cos2 a − sin2 b → simplify cos2 a − sin2 b = cos2 a − sin2 b ( ) 10. To determine all the solutions to the trigonometric equation 2 cos2 x + π2 = 1 such that 0 ≤ x ≤ 2π, it ( ) is necessary to ﬁrst calculate the value of cos x + π2 and then to apply the cosine identity for the sum of angles. 63 ( 2 cos2 x + ( 2 cos2 x + ( cos2 x + √ ( cos2 x + π) =1 2) π = 1 → ÷ both sides by 2 2 ) √ 1 π = → both sides 2 2 √ x) 1 = → rationalize denominator 2 2 ( √ ) (√ ) ( x) 2 1 cos x + = √ 2 2 2 √ ( x) 2 = √ → simplify cos x + 2 4 √ ( x) 2 cos x + = 2 2 Now apply cos(a + b) = cos a cos b − sin a sin b cos(a + b) = cos a cos b − sin a sin b ( π) π π cos x + = cos x cos − sin x sin 2 2 2 ( π) cos x + = cos x(0) − sin x(1) 2) ( π cos x + = − sin x 2 √ 2 sin x = → ÷ by − 1 2√ 2 sin x = − 2 The sine function √ is negative in the 3rd and 4th quadrants. Therefore, there are 2 angles that have the value 7π of sine equal to − 22 . These angles are 5π 4 and 4 . Sum and Diﬀerence Identities for Sine and Tangent Review Exercises: 1. To ﬁnd the exact value of presented here will use: 17π 12 , there is more than one combination that can be used. The solution ( ) 17π 14π 3π = sin + 12 12 12 ( ) 7π π 17π = sin + sin 12 6 4 sin and the sine identity for the sum of angles sin(a + b) = sin a cos b + cos a sin b, will be applied to determine the exact value. 64 sin(a + b) = sin a cos b + cos a sin b ( ) ( ) ( ) 7π ( 7π ( 7π π π) π) sin + = sin cos + cos sin 6 4 6 4 6 4 ( ) (√ ) ( ) ) ( )( 7π π 3 1 1 1 √ √ sin + → multiply + = − 6 4 2 2 2 2 √ ( ) 7π π 1 3 + sin = − √ − √ → simplify 6 4 2 2 2 2 √ ( ) −1 − 3 7π π √ sin + = → rationalize denominator 6 4 2 2 ( ) ( √ √ ) ( ) 7π π 2 −1 − 3 √ + sin = √ 6 4 2 2 2 √ √ ( ) − 2− 6 7π π √ sin + = → simplify 6 4 2 4 √ √ ( ) 7π π − 2− 6 √ sin + = 6 4 2 4 2. To determine the exact value of sin 345◦ , the sine identity for the sum of angles can be used to ﬁnd the sum of (300◦ + 45◦ ). sin(300◦ + 45◦ ) = sin 300◦ cos 45◦ − cos 300◦ sin 45◦ ( √ )( ) ( )( ) 3 1 1 1 ◦ ◦ √ sin(300 + 45 ) = − + −√ → multiply 2 2 2 2 √ 1 3 ◦ ◦ sin(300 + 45 ) = √ + √ → simplify 2 2 2 2 √ − 3+1 ◦ ◦ √ sin(300 + 45 ) = → rationalize denominator 2 2 (√ ) ( √ ) 2 − 3 + 1 √ sin(300◦ + 45◦ ) = √ 2 2 2 √ √ − 6+ 2 ◦ ◦ √ → simplify sin(300 + 45 ) = 2 4 √ √ − 6+ 2 sin(300◦ + 45◦ ) = 4 5 and is located in the 3rd quadrant and sin z = 45 and is located in the 2nd quadrant, the 3. If sin y = − 13 value of the adjacent side can be found by using the Pythagorean Theorem. Then the value of sin(y + z) can be determined by using the sine identity for the sum of angles. 65 (h)2 = (s1 )2 + (s2 )2 (13)2 = (−5)2 + (s2 )2 √ 169 = 25 + (s2 )2 √ 144 = s2 12 = s In the 3rd quadrant this value is negative. (h)2 = (s1 )2 + (s2 )2 (5)2 = (4)2 + (s2 )2 25 = 16 + (s2 )2 √ √ 9 = s2 3=s In the 2nd quadrant this value is negative. 66 sin(y + z) = sin y cos z + cos y sin z ( )( ) ( )( ) 5 3 12 4 sin(y + z) = − − + − → multiply 13 5 13 5 15 48 − → simplify sin(y + z) = 65 65 33 sin(y + z) = − 65 4. sin 25o cos 5o + cos 25o sin 5o is the expanded form of sin(a + b). Therefore the angle is sin(25o + 5o ) which equals sin30o and sin 30o = 12 . 5. To show that sin(a + b) · sin(a − b) = cos2 b − cos2 a, the sine identity for both the sum and the diﬀerence of angles must be used. Then the Pythagorean identity sin2 θ + cos2 θ = 1, or a form of this identity, will be applied. sin(a + b) · sin(a − b) = cos2 b − cos2 a (sin a cos b − cos a sin b)(sin a cos b + cos a sin b) = cos2 b − cos2 a → multiply sin2 a cos2 b − sin a sin b cos a cos b + sin a sin b cos a cos b − cos2 a sin2 b → simplify sin2 a cos2 b − cos2 a sin2 b = cos2 b − cos2 a → sin2 a = 1 − cos2 a → sin2 b = 1 − cos2 b (1 − cos2 a) cos2 b − cos2 a(1 − cos2 b) = cos2 b − cos2 a → expand cos2 b − cos2 a cos2 b − cos2 a + cos2 ba cos2 b = cos2 b − cos2 a → multiply cos2 b − cos2 a = cos2 b − cos2 a 6. To determine the value of tan(π + θ) the tangent identity for the sum of angles must be applied. This tan a+tan b identity is tan(a + b) = 1−tan a tan b tan a + tan b 1 − tan a tan b tan π + tan θ tan(π + θ) = 1 − tan π tan θ (0) + tan θ tan(π + θ) = 1 − (0) tan θ tan θ tan(π + θ) = 1 tan(a + b) = 7. To determine the exact value of tan 15o , the tangent identity for the diﬀerence of angles must be used since no special angles have a sum of 15o . However, 15o is the diﬀerence between 45o and 30o . Therefore, tan a−tan b tan(a − b) = 1+tan a tan b will be used. 67 tan a − tan b 1 + tan a tan b tan 45o − tan 30o tan(45o − 30o ) = 1 + tan 45o tan 30o 1 − √13 ( ) → simplify → common deno min ator tan(45o − 30o ) = 1 + 1 √13 tan(a − b) = tan(45 − 30 ) = o o tan(45o − 30o ) = √ √3 (1) − √1 3 3 √ → simplify 3 1 √ (1) + √ 3 3 √ 3−1 √ 3 √ → divide 3+1 √ 3 (√ tan(45 − 30 ) = o o 3−1 √ 3 )( √ ) 3 √ → simplify 3+1 √ 3−1 → rationalize deno min ator tan(45 − 30 ) = √ 3+1 ) (√ ) (√ 3−1 3−1 o o √ tan(45 − 30 ) = √ 3+1 3−1 √ √ 9−2 3+1 √ tan(45o − 30o ) = → simplify 9−1 √ 3−2 3+1 tan(45o − 30o ) = 3−1 √ 4 − 2 3 o o tan(45 − 30 ) = → reduce fraction 2√ tan(45o − 30o ) = 2 − 3 o o 8. To verify that sin π2 = 1 the sine identity for the sum of angles will be used. π ( π π) = sin + 2 4 4 sin(a + b) = sin a cos b + cos a sin b (π π ) π π π π sin + = sin cos + cos sin 4 4 4 4 4 4 (π π ) ( 1 ) ( 1 ) ( 1 ) ( 1 ) √ √ sin + = √ + √ → multiply 4 4 2 2 2 2 (π π ) 1 1 sin + = √ + √ → simplify 4 4 4 4 (π π ) 1 1 sin + = + 4 4) 2 2 (π π + =1 sin 4 4 sin 9. To reduce cos(x + y) cos y + sin(x + y) sin y to a single term requires the use of the cosine identity for the sum of angles and the sine identity for the sum of angles. 68 cos(x + y) cos y + sin(x + y) sin y → expand cos x cos2 y − sin x sin y cos y + sin x sin y cos y + cos x sin2 y → simplify cos x cos2 y + cos x sin2 y → remove common factor (cos x) cos x(cos2 y + sin2 y) → sin2 x + cos2 x = 1 cos x(1) = cos x cos(x + y) cos y + sin(x + y) sin y = cos x ) ( ) ( 10. To solve 2 tan2 x + π6 − 1 = 7 for all values in the interval [0, 2π), the value of tan π6 must be determined and then the tangent identity for the sum of angles must be applied to ﬁnd the values within the stated interval. 69 ( π) 2 tan2 x + −1=7 ( ) 6 π 2 tan2 x + −1+1=7+1 6( π) 2 tan2 x + = 8 → ÷ both sides by 2 6) ( π √ = 4 → both sides tan2 x + 6 √ ( π) √ tan2 x + = 4 → simplify 6 ) ( π =2 tan x + 6 π tan x + tan 6 =2 1 − tan x tan π6 π π tan x + tan = 2(1 − tan x tan ) 6 6 1 1 tan x + √ = 2(1 − tan x √ ) → rationalize deno min ator 3 3 (√ ) ( (√ ) ( ) ) 3 1 3 tan x √ √ tan x + √ =2− √ → multiply 3 3 3 3 √ √ 3 3 tan x √ → simplify tan x + √ = 2 − 9 9 √ √ 3 3 tan x √ tan x + √ = 2 − 9 9 √ √ 3 3 tan x + = 2 − √ → common deno min ator (LS) 3 3 √ 3 tan x + 3 tan x ≈ 1.4226 3 1.5774 tan x ≈ 1.4226 1.4226 1.5774 tan x ≈ 1.5774 1.5774 tan x ≈ 0.9019 tan−1 (tan x) ≈ tan−1 (0.9019) x ≈ 0.7338rad To determine the values, change the radians to degrees. The angles will be located in the 1st and 3rd quadrants. ( o) 0.7338 180 ≈ 42o . The angle in the 3rd quadrant would be approximately 222o . π Double-Angle Identities Review Exercises: Pages 260 – 265 1. If sin x = 45 , and is the 2nd quadrant then 70 (h)2 = (s1 )2 + (s2 )2 (5)2 = (4)2 + (s2 )2 25 = 16 + (s2 )2 √ √ 9 = s2 3=s In the 2nd quadrant, this value is negative. For the above angle in standard position, cos x = − 35 and x = − 34 . The double-angle identities will be used to determine the exact values of sin 2x, cos 2x, tan 2x. 71 sin 2x = 2 sin x cos x ( )( ) 4 3 sin 2x = 2 − → multiply 5 5 24 sin 2x = − 25 cos 2x = cos2 x − sin2 x )2 ( )2 ( 3 4 − cos 2x = − 5 5 9 16 cos 2x = − 25 25 7 cos 2x = − 25 2 tan x tan 2x = 1 − tan2 x ( ) 2 43 tan 2x = ( )2 → simplify 1 − − 43 tan 2x = −8 3 −7 9 ( → simplify −8 3 24 tan 2x = 7 )( tan 2x = −9 7 ) 2. cos2 15◦ − sin2 15◦ is the identity for cos 2a. cos 2a = cos2 a − sin2 a If a = 15◦ than 2a = 2(15◦ ) = 30◦ √ ◦ cos 30 = 3 2 3. Verify: cos 3θ = 4 cos3 θ − 3 cos θ. To verify this identity, the cosine identity for the sum of angles and the double-angle identities for cosine and sine will have to be applied. 72 cos 3θ = 4 cos3 θ − 3 cos θ cos(a + b) = cos a cos b − sin a sin b → a = 2θ; b = θ cos(2θ + θ) = (2 cos2 θ − 1) cos θ − (2 sin θ cos θ) sin θ → expand cos(2θ + θ) = 2 cos3 θ − cos θ − 2 sin2 θ cos θ cos(2θ + θ) = cos θ(2 cos2 θ − 1 − 2 sin2 θ) → sin2 θ + cos2 θ = 1 cos(2θ + θ) = cos θ(2 cos2 θ − 1 − 2(1 − cos2 θ)) → simplify cos(2θ + θ) = cos θ(2 cos2 θ − 1 − 2 + 2 cos2 θ) → simplify cos(2θ + θ) = cos θ(4 cos2 θ − 3) → expand cos(2θ + θ) = 4 cos3 θ − 3 cos θ 4. Verify: sin 2t − tan t = tan t cos 2t. To verify this identity, the quotient identity for tangent must be used as well as the double-angle identities for sine and cosine. sin 2t − tan t = tan t cos 2t → sin 2t = 2 sin t cos t sin t → tan t = cos t sin t 2 sin t cos t − = tan t cos 2t → common denominator cos t ( ) cos t sin t 2 sin t cos t − = tan t cos 2t → simplify cos t cos t 2 sin t cos2 t sin t − = tan t cos 2t → simplify cos t cos t 2 sin t cos2 t − sin t = tan t cos 2t → common factor (sin t) cos t (sin t)2 cos2 t − 1 = tan t cos 2t → cos 2t = 2 cos2 −1 cos t (sin t) cos 2t = tan t cos 2t → cos 2t = 2 cos2 t − 1 cos t (sin t) cos 2t sin t = tan t cos 2t → tan t = cos t cos t tan t cos 2t = tan t cos 2t 9 5. If sin x = − 41 and is located in the 3rd quadrant, then: 73 (h)2 = (s1 )2 + (s2 )2 (41)2 = (−9)2 + (s2 )2 √ 1681 = 81 + (s2 )2 √ 1600 = s2 40 = s In the 3rd quadrant this value is negative. For the above angle in standard position, cos x = − 40 41 and tan x = used to determine the exact values of sin 2x, cos 2x, tan 2x. 9 40 . The double-angle identities will be sin 2x = 2 sin x cos x ( )( ) 9 40 sin 2x = 2 − − → multiply 41 41 720 sin 2x = 1681 cos 2x = 2 cos2 x − 1 ( )2 40 cos 2x = 2 − −1 41 3200 cos 2x = − 1 → common denominator 1681 ( ) 3200 1681 cos 2x = −1 → simplify 1681 1681 1519 cos 2x = 1681 sin 2x 720 1519 tan 2x = → sin 2x = ; cos 2x = cos 2x 1681 1681 tan 2x = 720 1681 1591 1681 ( → simplify 720 1681 720 tan 2x = 1519 tan 2x = )( 1681 1519 ) 6. To ﬁnd all the solutions for x in the equation sin 2x + sin x = 0 such that 0 ≤ x < 2π, the double-angle identity for sine must be used. 74 sin 2x + sin x = 0 → sin 2x = 2 sin x cos x 2 sin x cos x + sin x = 0 → common factor (sin x) (sin x)2 cos x + 1 = 0 Then sin x = 0 or 2 cos x + 1 = 0 → solve sin−1 (sin x) = sin−1 (0) x = 0, π 2 cos x + 1 = 0 cos x = − −1 cos 1 2 −1 ( (cos x) = cos x= 1 − 2 ) 2π 4π , 3 3 7. To ﬁnd all the solutions for x in the equation cos2 x − cos 2x = 0 such that 0 ≤ x < 2π, the double-angle identity for cosine must be used. cos2 x − cos 2x = 0 → cos 2x = 2 cos2 x − 1 cos2 x − (2 cos2 x − 1) = 0 → simplify − cos2 x + 1 = 0 → ÷ by − 1 cos2 x − 1 = 0 → factor (cos x + 1)(cos x − 1) = 0 → solve Then cos x + 1 = 0 or cos x − 1 = 0 cos x = −1 −1 cos cos x = 1 −1 (cos x) = cos x=π cos−1 (cos x) = cos−1 (1) x=0 (−1) 8. The formula for cos2 x in terms of the ﬁrst power of cosine is cos2 x = 21 (cos 2x + 1). To express cos4 x in terms of the ﬁrst power of cosine, the ﬁrst step is to realize that cos4 x = (cos2 x)2 . Therefore: 75 1 (cos 2x + 1) → square both sides 2 ]2 [ 1 2 2 (cos x) = (cos 2x + 1) → expand 2 1 cos2 x = (cos2 2x + 2 cos 2x + 1) 4 1 1 if cos2 x = (cos 2x + 1) → replace x with 2x then cos2 2x = (cos 4x + 1) 2[ 2 ] 1 1 cos4 x = (cos 4x + 1) + 2 cos 2x + 1 → expand 4 2 [ ] 1 1 1 4 cos x = cos 4x + + 2 cos 2x + 1 → simplify 4 2 2 [ ] 1 1 3 4 cos x = cos 4x + 2 cos 2x + → multiply 4 2 2 2 3 1 cos4 x = cos 4x + cos 2x + → common denominator 8 4 8 ( ) 1 2 2 3 cos4 x = cos 4x + cos 2x + 8 2 4 8 1 4 3 4 cos x = cos 4x + cos 2x + → simplify 8 8 8 cos 4x + 4 cos 2x + 3 4 cos x = 8 cos2 x = 9. The formula for sin2 x in terms of the ﬁrst power of cosine is sin2 x = 12 (1 − cos 2x). To express sin4 x in terms of the ﬁrst power of cosine, the ﬁrst step is to realize that sin4 x = (sin2 x)2 . Therefore: 76 1 (1 − cos 2x) → square both sides 2 [ ]2 1 (sin2 x)2 = (1 − cos 2x) → expand 2 1 sin2 x = (1 − 2 cos 2x + cos2 2x) 4 1 1 if sin2 x = (1 − cos 2x) → replace x with 2x then cos2 2x = (cos 4x + 1) 2[ 2 ] 1 1 sin4 x = 1 − 2 cos 2x + (cos 4x + 1) → expand 4 2 [ ] 1 1 1 sin4 x = 1 − 2 cos 2x + cos 4x + → simplify 4 2 2 [ ] 1 3 1 sin4 x = − 2 cos 2x + cos 4x → multiply 4 2 2 3 2 1 sin4 x = − cos 2x + cos 4x → common denominator 8 ( 4 ) 8 3 2 2 1 sin4 x = − cos 2x + cos 4x 8 2 4 8 3 4 1 sin4 x = − cos 2x + cos 4x → simplify 8 8 8 3 − 4 cos 2x + cos 4x sin4 x = 8 sin2 x = 10. a) To rewrite sin2 x cos2 2x in terms of the ﬁrst power of cosine, determine the product by using the formulas: sin2 x = 12 (1 − cos2x) and cos2 2x = 21 (cos 4x + 1) sin2 x cos2 2x [ ][ ] 1 1 (1 − cos 2x) (cos 4x + 1) → expand 2 2 ( )( ) 1 1 1 1 − cos 2x cos 4x + → expand 2 2 2 2 ( ) 1 1 1 1 1 cos 4x + − cos 2x cos 4x − cos 2x → common factor 4 4 4 4 4 1 (cos 4x + 1 − cos 2x cos 4x − cos 2x) → rearrange 4 1 (1 − cos 2x + cos 4x − cos 2x cos 4x) 4 b) To rewrite tan4 2x in terms of the ﬁrst power of cosine, the quotient identity for tangent will be used along 4x . with cos4 x = cos 4x+48cos 2x+3 and sin4 x = 3−4 cos 2x+cos 8 77 sin4 x cos4 x sin4 2x tan4 2x = cos4 2x tan4 x = tan4 x = tan4 x = 3−4 cos 2x+cos 4x 8 3+4 cos 2x+cos 4x 8 3−4 cos 4x+cos 8x 8 3+4 cos 4x+cos 8x 8 → replace x with 2x → simplify ( )( ) 3 − 4 cos 4x + cos 8x 8 tan 2x = → multiply 8 3 + 4 cos 4x + cos 8x ) ( 3 − 4 cos 4x + cos 8x tan4 2x = 3 + 4 cos 4x + cos 8x 4 Half-Angle Identities Review Exercises: 1. To determine the exact value of cos 112.5◦ , the angle must √ be expressed in the form of a half-angle. Once θ this is done, the half-angle identity for cosine, cos 2 = ± cos 2θ+1 can be used to determine the exact value. 78 225◦ cos 112.5◦ = cos √ 2 θ cos θ + 1 cos = ± 2 2 √ ◦ 225 cos 225◦ + 1 cos =± 2 2 √ −1 √ +1 225◦ 2 cos =± → (common denominator) 2 2 v (√ ) u u √1 + √2 1 t 2 225◦ 2 cos =± → simplify 2 2 √ √ −1 √ √2 + 225◦ 2 2 cos =± → simplify 2 2 √ √ −1+ 2 √ 225◦ 2 cos =± → simplify 2 2 v( u √ ) u −1 + 2 ( 1 ) 225◦ t √ cos =± → simplify 2 2 2 v( u √ ) u −1 + 2 225◦ t √ cos =± → rationalize denominator 2 2 2 v( u √ ) (√ ) u −1 + 2 2 225◦ t √ √ → simplify cos =± 2 2 2 2 v( u √ √ ) u − 2+ 4 225◦ t √ cos → simplify =± 2 2 4 v( ) u √ u − 2+2 225◦ t cos → simplify =± 2 4 225◦ cos =± 2 √ √ − 2+2 2 112.5◦ is an angle located in the 2nd quadrant. The cosine of an angle in this quadrant is negative. The exact value of this angle is: 225◦ =− cos 2 √ √ − 2+2 2 2. To determine the exact value of 105◦ , the angle√must be expressed in the form of a half-angle. Once this θ is done, the half-angle identity for sine, sin θ2 = ± 1−cos can be used to determine the exact value. 2 79 210◦ sin 105◦ = sin √ 2 θ 1 − cos θ sin = ± 2 2 √ 210◦ 1 − cos 210◦ sin =± 2 2 v ( √ ) u u1 − − 3 t 2 210◦ sin =± → common denominator 2 2 v u ( ) ( √ ) u1 2 − − 3 t 2 2 210◦ sin =± → simplify 2 2 v (√ ) u u2 + 3 t2 2 210◦ sin =± → simplify 2 2 √ √ 2+ 3 210◦ 2 sin =± → simplify 2 2 v( u √ ) u 2 + 3 (1) 210◦ t sin =± → simplify 2 2 2 v( u √ ) ◦ u 2+ 3 210 sin = ±t → simplify 2 2 210◦ sin =± 2 √ √ 2+ 3 2 105◦ is an angle located in the 2nd quadrant. The sine of an angle in this quadrant is positive. The exact value of this angle is: 210◦ sin = 2 √ 2+ 2 √ 3 be expressed in the form of a half-angle. Once this 3. To determine the exact value of tan 7π 8 , the angle must √ θ 1−cos θ is done, the half-angle identity for tangent, tan 2 = ± 1+cos θ can be used to determine the exact value. 80 7π 7π = tan 4 8 √2 θ 1 − cos θ tan = ± 2 1 + cos θ √ 7π 1 − cos 7π 4 tan 4 = ± 2 1 + cos 7π 4 v u 7π u 1 − √12 tan 4 = ±t → common denominator 2 1 + √12 v(√ ) u 2 u √ 1 − √1 7π u 2 2 4 = ±t ( √ ) tan → simplify 2 1 2 √ 1 + √2 2 v(√ ) u 2 u √ − √1 7π u 2 2 4 = ±t ( √ ) → simplify tan 2 1 2 √ √ + 2 2 v(√ u 2−1 ) u √ 7π u 2 ) → simplify tan 4 = ±t ( √ 2+1 2 √ tan 2 v( )( √ ) u √ 7π u 2 − 1 2 √ √ → simplify tan 4 = ±t 2 2 2+1 v( u √ √ ) 7π u 4 − 2 √ tan 4 = ±t √ → simplify 2 4+ 2 v( u √ ) √ )( 7π u 2− 2 2− 2 4 t √ √ tan =± → rationalize denominator 2 2+ 2 2− 2 v( u √ √ ) u 4−4 2+ 4 t √ tan =± → simplify 2 4− 4 v( ) u √ 7π 7π u 4−4 2+2 4 t tan =± → simplify tan 4 2 4−2 2 √ 7π √ tan 4 = ± 3 − 2 2 2 7π 4 v( u √ ) u 6−4 2 t =± → simplify 2 is an angle located in the 2nd quadrant. The tangent of an angle in this quadrant is negative. The exact value of this angle is: 7π 8 81 tan 7π 4 2 √ √ =± 3−2 2 4. To determine the exact value of tan π8 , the angle must be expressed in the form of a half-angle. Once this √ θ is done, the half-angle identity for tangent, tan θ2 = ± 1−cos 1+cos θ can be used to determine the exact value. π π = tan 4 8 √2 θ 1 − cos θ tan = ± 2 1 + cos θ √ π 1 − cos π4 tan 4 = ± 2 1 + cos π4 v u π u 1 − √12 4 → common denominator tan = ±t 2 1 + √12 v(√ ) u 2 u √ 1 − √1 π u 2 2 4 tan = ±t ( √ ) → simplify 2 √2 1 − √1 2 2 v√ u 2−1 u √ π u 2 4 t √ → simplify tan = ± 2+1 2 √ tan 2 tan tan tan tan tan π 4 2 π 4 2 π 4 2 π 4 2 π 4 2 v( )( √ ) u √ u 2−1 2 t √ √ =± → simplify 2 2+1 v( u √ √ ) u 4 − 2 √ = ±t √ → simplify 4+ 2 v( u √ ) u 2− 2 √ = ±t → rationalize denominator 2+ 2 v( u √ )( √ ) u 2− 2 2 − 2 √ √ = ±t → simplify 2+ 2 2− 2 v( u √ √ ) u 4−4 2+ 4 √ = ±t → simplify 4− 4 82 v( ) u √ u 4−4 2+2 t → simplify tan = ± 2 4−2 v( u √ ) π u 6−4 2 4 t tan = ± → simplify 2 2 √ π √ 4 tan = ± 3 − 2 2 2 π 4 is an angle located in the 1st quadrant. The tangent of an angle in this quadrant is positive. The exact value of this angle is: is an angle located in the 1st quadrant. The tangent of an angle in this quadrant is positive. The exact value of this angle is: π 8 tan 5. If sin θ = 7 25 π 4 2 √ √ =± 3−2 2 and is located in the 2nd quadrant, then: (h)2 = (s1 )2 + (s2 )2 (25)2 = (7)2 + (s2 )2 625 = 49 + (s2 )2 √ √ 576 = s2 24 = s In the 2nd quadrant this value is negative and cos θ = − 24 25 . Table 3.1 Steps = cos θ2 √ ± cos 2θ+1 sin θ2 √ θ ± 1−cos 2 83 tan θ2 √ θ ± 1−cos 1+cos θ Table 3.1: (continued) Steps cos θ = − 24 25 simplify Common denominator simplify simplify simplify √ √ 25 · 2 = 5 2 Rationalize denominator 2nd quadrant angle sin θ2 √ 1−(− 24 25 ) ± 2 √ 24 1+ 25 ± 2 √ 25 24 + ± 25 2 25 √( ) ( ) 49 1 ± 25 2 √ 49 ± 50 7 ± 5√ 2 7 ± 5√ 2 cos θ2 √ 24 (− 25 )+1 ± 2 √ 24 − 25 +1 ± 2 √ 24 25 + ± − 25 2 25 √( ) ( ) 1 1 ± 25 2 √ 1 ± 50 (√ ) sin θ2 = 2 2 = 1 ± 5√ 2 √ ± 7102 1 ± 5√ 2 √ 7 2 10 tan θ2 √ 24 1−(− 25 ) ± 1+ − 24 ( 25 ) √ 1+ 24 ± 1− 25 24 √ 2425 25 24 25 + − ± − 25 2 25 25 2 25 √ √( ) ( ) 49 49 25 ± 25 1 = ± 25 1 √ 25 49 ± 1 (√ ) 2 2 √ cos θ2 = − 102 = √ ± 102 ± 17 ± 17 tan θ2 = −7 sec b 6. To verify the identity tan 2b = sec b csc b+csc b , the half-angle identity for tangent must be used as well as the reciprocal identities for secant and cosecant. 84 sec b b = → common factor (csc b) 2 sec b csc b + csc b b sec b tan = → reciprocal identities 2 csc b(sec b + 1) tan 1 b = 2 1 sin b tan b = 2 1 sin b cos b tan b = 2 tan b = 2 tan ( cos1 b cos b 1 cos b ) → multiply +1 → common denominator 1 sin b 1 cos b ( cos b ) 1 → simplify 1 sin b cos b + cos b sin b 1 cos b → simplify 1+cos b sin b cos b + ( )( ) b 1 sin b cos b = → simplify 2 cos b 1 + cos b ( )( ) b 1 sin b cos b tan = → simplify 2 cos b 1 + cos b √ b sin b b 1 − cos b tan = and tan = ± → half - angle identity for tan gent 2 1 + cos b 2 1 + cos b √ 1 − cos b sin b ∴± = → square both sides 1 + cos b 1 + cos b )2 ( ( √ )2 1 − cos b sin b = → expand ∴ ± 1 + cos b 1 + cos b tan 1 − cos b sin2 b = 1 + cos b (1 + cos b)2 (1 − cos b)(1 + cos b)(1 + cos b) = sin2 b(1 + cos b) → common factor (1 + cos b) (1 − cos b)(1 + cos b) (1 + cos b) sin2 b = → simplify (1 + cos b) (1 + cos b) (1 − cos b)(1 + cos b) = sin2 b → multiply 1 − cos2 b = sin2 b → sin2 b + cos2 b = 1 sin2 b = sin2 b sin c 7. To verify the identity cot 2c = 1−cos c , the quotient identity for cotangent must be applied as well as the half-angle identities for sine and cosine. 85 sin c cos θ c = → cot θ = 2 1 − cos c sin θ cos 2c sin c = → half - angle identities sin 2c 1 − cos c cot √ ± cos2c+1 sin c √ = → simplify (LS) 1 − cos c 1−cos c ± 2 √( )( ) cos c + 1 2 sin c → simplify ± = 2 1 − cos c 1 − cos c √( )( ) 2 sin c cos c + 1 ± = → square both sides 1 − cos c 1 − cos c 2 ( √ )2 ( sin c )2 → simplify ± f raccos c + 11 − cos c = 1 − cos c cos c + 1 sin2 c = → expand 1 − cos c (1 − cos c)2 (cos c + 1)(1 − cos c)(1 − cos c) = sin2 c(1 − cos c) → common factor sin2 c (1 − cos c) (cos c + 1)(1 − cos c) (1 − cos c) = → simplify (1 − cos c) (1 − cos c) (cos c + 1)(1 − cos c) = sin2 c → multiply 1 − cos2 c = sin2 c → sin2 c + cos2 c = 1 sin2 c = sin2 c 8 8. If sin u = − 13 , the angle must be located in either the 3rd or 4th quadrant since the sine function is negative here. (h)2 = (s1 )2 + (s2 )2 (13)2 = (−8)2 + (s2 )2 √ √ 169 = 64 + (s2 )2 √ 105 = s2 105 = s √ − 105 is inadmissible in the half-angle formula. Therefore the angle is in the 4th quadrant and cos u = 86 √ 105 13 √ u cos u + 1 cos = ± 2 2 √√ 105 u 13 + 1 cos = ± → common denominator 2 2 √√ ( 13 ) 105 u 13 + 13 1 cos = ± → simplify 2 2 √√ 105+13 u 13 cos = ± → simplify 2 2 v( )( ) u √ u u 105 + 13 1 t cos = ± → simplify 2 13 2 √√ u 105 + 13 cos = 2 26 The angle is located in the 4th quadrant where the cosine function has a positive value. 9. To solve the trigonometric equation 2 cos2 identity for cosine must be used. x 2 = 1 for values of x such that 0 ≤ x < 2π the half-angle x = 1 → ÷ both sides by 2 2 x 1 cos2 = → half - angle identity 2 2 )2 ( √ 1 cos x + 1 = → simplify ± 2 2 2 cos2 cos x + 1 1 = → simplify 2 2 2(cos x + 1) = 2 → multiply 2 cos x + 2 = 2 → solve 2 cos x + 2 − 2 = 2 − 2 2 cos x 0 = 2 2 cos x = 0 cos−1 (cos x) = cos−1 (0) π 3π x = and 2 2 10. To solve the trigonometric equation tan a2 = 4 for all values of x such that 0◦ ≤ x < 360◦ , the half-angle identity for tangent must be used. 87 ( √ tan a = 4 → half - angle identity 2 ) 1 − cos a = 4 → square both sides 1 + cos a ( √ )2 1 − cos a = (4)2 → simplify ± 1 + cos a ± 1 − cos a = 16 1 + cos a 16(1 + cos a) = 1 − cos a → multiply 16 + 16 cos a = 1 − cos a → solve 16 − 16 + 16 cos a = 1 − cos a − 16 16 cos a = − cos a − 15 16 cos a + cos a = − cos a + cos a − 15 17 cos a −15 = 17 17 15 cos a = − → use calculator 17 ( ) 15 −1 −1 cos (cos a) = cos − 17 The cosine function has a negative value in both the 2nd and 3rd quadrants. There are 2 values for angle a. a ≈ 152◦ and a ≈ 108◦ Product-and Sum, Sum-and-Product and Linear Combinations of Identities Review Exercises: 1. To express sin 9x + sin 5x as a product, the sum to product formula for sine must be used. ( ) ( ) α+β α−β · cos → α = 9x 2 2 ( ) ( ) 9x + 5x 9x − 5x sin 9x + sin 5x = 2 sin · cos → simplify 2 2 sin α + sin β = 2 sin sin 9x + sin 5x = 2 sin(7x) · cos(2x) 2. To express cos 4y − cos 3y as a product, the diﬀerence to product formula for cosine must be used. 88 ( cos α − cos β = −2 sin α+β 2 ) ( ( · sin α−β 2 ) ( 4y + 3y · sin 2 ( ) (y) 7y cos 4y − cos 3y = −2 sin · sin 2 2 cos 4y − cos 3y = −2 sin ) → α = 4y → β = 3y ) 4y − 3y → simplify 2 3a−cos 5a 3. To verify cos sin3a+sin 5a = − tan(−a), the diﬀerence to product formula for cosine and the sum to product formula for sine must be used. In addition, the quotient identity for tangent must be applied. ( cos α − cos β = −2 sin ( cos 3a − cos 5a = −2 sin α+β 2 ) 3a + 5a 2 ( · sin α+β 2 ) ( · sin ) → α = 3a → β = 5a ) 3a + 5a → simplify 2 cos 3a − cos 5a = −2 sin 4a · sin(−a) ( ) ( ) α+β α+β sin α − sin β = 2 sin · cos → α = 3a 2 2 ( ) ( ) 3a + 5a 3a + 5a sin 3a − sin 5a = 2 sin · cos → simplify 2 2 sin 3a − sin 5a = 2 sin 4a · cos(−a) cos 3a − cos 5a = − tan(−a) → substitute above solutions sin 3a + sin 5a −2 sin 4a · sin(−a) = − tan(−a) → simplify 2 sin 4a · cos(−a) ( · sin(−a) (4a −2(sin ( · cos(−a) = − tan(−a) → simplify 2 sin 4a sin(−a) sin θ − = − tan(−a) → tan θ = → θ = −a cos(−a) cos θ sin(−a) sin(−a) − =− cos(−a) cos(−a) sin(−a) − = − tan(−a) cos(−a) 4. To express the product sin(6θ) sin(4θ) as a sum, the product to sum formula for sine must be used. 1 [cos(α − β) − cos(α + β)] → α = 6θ 2 → β = 4θ 1 sin(6θ) sin(4θ) = [cos(6θ − 4θ) − cos(6θ + 4θ)] → simplify 2 1 sin(6θ) sin(4θ) = [cos(2θ) − cos(10θ)] 2 sin α sin β = 89 5. a) To express 5 cos x − 5 sin x as a linear combination the formula a cos x + b sin x = C cos(x − d) must be used. From the above, a = 5 and b = −5. This indicates that the angle is located in the 4th quadrant. The Pythagorean Theorem can be used to determine the value of C. a2 + b2 = c2 (5)2 + (−5)2 = c2 → simplify √ √ √ 50 = c2 → both sides √ √ 25 · 2 = c → simplify ( 50) √ 5 2=c adj cos d = hyp 5 cos d = √ → rationalize deno minator 5 2 (√ ) 5 2 √ cos d = √ → simplify 5 2 2 √ √ √ 5 5 5 2 2 cos d = √ = = 10 2 5 4 In the 4th quadrant d has a value of 7π 4 radians (unit circle) a cos x + b sin x = C cos(x − d) ( ) √ 7π 5 cos x − 5 sin x = 5 5 cos x − 4 b) To express −15 cos 3x − 8 sin 3x as a linear combination, the formula a cos x + b sin x = C cos(x − d) must be used. From the above, a = −15 and b = −8. This indicates that the angle is located in the 3rd quadrant. The Pythagorean Theorem can be used to determine the value of C. a2 + b2 = c2 (−15)2 + (−8)2 = c2 → simplify √ √ √ 289 = c2 → both sides 17 = c adj cos d = hyp 15 cos d = − 17 ( ) 15 cos−1 (cos d) = cos−1 17 d ≈ 28◦ The angle has already been determined to be in the 4th quadrant. Therefore an angle of 28◦ in standard position in the this quadrant would have a value of approximately 208◦ or 3.63 radians. 90 a cos x + b sin x = C cos(x − d) −15 cos 3x − 8 sin 3x = 17 cos(x − 208◦ ) −15 cos 3x − 8 sin 3x = 17 cos(x − 3.63 rad) 6. To solve the equation sin 4x + sin 2x = 0 for all values of x such that 0 ≤ x < 2π, the sum to product formula for sine must be used. ( sin α + sin β = 2 sin ( sin 4x + sin 2x = 2 sin α+β 2 ) 4x + 2x 2 ( · cos α−β 2 ) ( · cos ) → α = 4x → β = 2x ) 4x − 2x → simplify 2 sin 4x + sin 2x = 2(sin 3x · cos x) 2(sin 3x · cos x) = 0 → solve 2(sin 3x · cos x) 0 = 2 2 sin 3x · cos x = 0 Then sin 3x = 0 Or cos x = 0 sin 3x = 0 The interval 0 ≤ x < 2π will be tripled since the equation deals with sin 3x. This will give the results in the interval 0 ≤ x < 6π 3x = 0, π, 2π, 3π, 4π, 5π To obtain the values of x, each of the above answers must be divided by 3. π 2π 3π 4π 5π , , , , 3 3 3 3 3 π 2π 4π 5π x = 0, , , π, , 3 3 3 3 cos x = 0 π 3π x= , 2 2 x = 0, When sin 4x + sin 2x = 0 is solved for all values of x such that 0 ≤ x < 2π, the results are: x = 0, 4π 3π 5π π π 2π , , , π, , , 3 2 3 3 2 3 7. To solve the equation cos 4x + cos 2x = 0 for all values of x such that 0 ≤ x < 2π, the sum to product formula for cosine must be used. 91 ( cos α + cos β = 2 cos ( cos 4x + cos 2x = 2 cos α+β 2 ) 4x + 2x 2 ( · cos α−β 2 ) ( · cos ) → α = 4x → β = 2x ) 4x − 2x → simplify 2 cos 4x + cos 2x = 2 cos 3x · cos x 2 cos 3x · cos x = 0 → solve 0 2 cos 3x) · cos x = 2 2 cos 3x · cos x = 0 Then cos 3x = 0 Or cos x = 0 cos 3x = 0 The interval 0 ≤ x < 2π will be tripled since the equation deals with cos 3x. This will give the results in the interval 0 ≤ x < 6π 3x = π 3π 5π 7π 9π 11π , , , , , 2 2 2 2 2 2 To obtain the values of x, each of the above answers must be divided by 3. π 3π 5π 7π 9π 11π , , , , , 6 6 6 6 6 6 π π 5π 7π 3π 11π x= , , , , , 6 2 6 6 2 6 cos x = 0 π 3π x= , 2 2 x= When cos 4x + cos 2x = 0 is solved for all values of x such that 0 ≤ x < 2π, the results are: x= π π 5π 7π 3π 11π , , , , , 6 2 6 6 2 6 8. To solve the equation sin 5x + sin x = sin 3x for all values of x such that 0 ≤ x < 2π, the sum to product formula for sine or the diﬀerence to product formula for sine must be used. The formula that is used depends upon how the equation is manipulated. However, the solution will not be aﬀected by the formula. 92 sin 5x + sin x = sin 3x → set = 0 sin 5x − sin 3x + sin x = sin 3x − sin 3x sin 5x − sin 3x + sin x = 0 → diﬀerence to product ( ) ( ) α−β α+β sin α + sin β = 2 sin · cos → α = 5x 2 2 ( sin 5x − sin 3x = 2 sin 5x − 3x 2 ) ( · cos → β = 3x ) 5x + 3x → simplify 2 sin 5x − sin 3x = 2 sin x · cos x 2 sin x · cos 4x + sin x = 0 → common factor sin x(2 cos 4x + 1) = 0 Then sin x = 0 Or 2 cos 4x + 1 = 0 sin 3x = 0 2 cos 4x + 1 = 0 2 cos 4x + 1 − 1 = 0 − 1 2 cos 4x = −1 2 cos 4x −1 = 2 2 1 cos 4x = − 2 x = 0, π The interval 0 ≤ x < 2π will be multiplied by 4 since the equation deals with cos 4x. This will give the results in the interval 0 ≤ x < 8π 4x = 2π 4π 8π 10π 14π 16π 20π 22π , , , , , , , 3 3 3 3 3 3 3 3 To obtain the values of x, each of the above answers must be divided by 4. 2π 4π 8π 10π 14π 16π 20π 22π , , , , , , , 12 12 12 12 12 12 12 12 π π 2π 5π 7π 4π 5π 11π x= , , , , , , , 6 3 3 6 6 3 3 6 x= When sin 5x + sin x = sin 3x is solved for all values of x such that 0 ≤ x < 2π, the results are: x = 0, 7π 4π 5π 11π π π 2π 5π , , , , π, , , , 6 3 3 6 6 3 3 6 9. The sum to product formula for sine will be used to simplify the equation f (t) = sin(200t+π)+sin(200t−π) 93 ( sin α + sin β = 2 sin α+β 2 ) ( · cos α−β 2 ) → α = 220t + π → β = 220t − π ) ( ) (220t + π) + (220t − π) (220t + π) − (220t − π) sin(220t + π) + sin(220t − π) = 2 sin · cos → simply 2 2 ( ) ( ) 400t 2π sin(220t + π) + sin(220t − π) = 2 sin · cos → simply 2 2 ( sin(220t + π) + sin(220t − π) = 2 sin 200t · cos π → cos π = −1 sin(220t + π) + sin(220t − π) = 2 sin 200t(−1) sin(220t + π) + sin(220t − π) = −2 sin 200t 10. To determine a formula for tan 4x the sum formula for tangent and the double- angle formula for tangent will be used. 94 tan 4x = tan(2x + 2x) → sum formula (tan gent) tan a + tan b → a = 2x tan(a + b) = 1 − tan a tan b → b = 2x tan 2x + tan 2x tan(2x + 2x) = → simplify 1 − tan 2x tan 2x 2 tan 2x → double angle formula tan(2x + 2x) = 1 − tan2 2x ( ) 2 tan x 2 1−tan 2x tan(2x + 2x) = ( )2 → simplify 2 tan x 1 − 1−tan 2x ( ) 4 tan x 1−tan2 x tan(2x + 2x) = 1− tan(2x + 2x) = tan(2x + 2x) = ( (2 tan x) (1−tan2 x) ( 2 )2 → common deno min tor 4 tan x 1−tan2 x ) 2 x) 4tan x 1 (1−tan (1−tan2 x)2 − (1−tan2 x)2 ) ( 2 4 tan x 1−tan2 x (1−tan2 x)2 −4 tan2 x (1−tan2 x)2 → simplify → simplify ( ) 4 tan x (1 − tan2 x)2 − 4 tan2 x ÷ → simplify 2 1 − tan x (1 − tan2 x)2 ( ) 4 tan x (1 − tan2 x)2 tan(2x + 2x) = · → simplify 2 1 − tan x (1 − tan2 x)2 − 4 tan2 x ( ) 2 4 tan x (1 − tan x)(1 − tan2 x) tan(2x + 2x) = · → simplify 2 (1 − tan2 x)2 − 4 tan2 x tan 1− x 4 tan x(1 − tan2 x) tan(2x + 2x) = → expand (1 − tan2 x)2 − 4 tan2 x 4 tan x − 4 tan3 x tan(2x + 2x) = → simplify 1 − 2 tan2 x + tan4 x − 4 tan2 x 4 tan x − 4 tan3 x tan(2x + 2x) = 1 − 6 tan2 x + tan4 x 4 tan x − 4 tan3 x tan(4x) = 1 − 6 tan2 x + tan4 x tan(2x + 2x) = Chapter Review Review Exercises: Pages 280 – 285 1. To determine the sine, cosine and tangent of an angle that has (−8, 15) on its terminal side, sketch the angle in standard position in the2nd quadrant. Use the Pythagorean Theorem to determine the length of the hypotenuse. 95 (h)2 = (s1 )2 + (s2 )2 (h)2 = (15)2 + (−8)2 (h)2 = 225 + 64 √ √ h2 = 289 h = 17 opp hyp 15 sin θ = 17 sin θ = adj hyp 8 cos θ = − 17 cos θ = √ opp adj 15 tan θ = − 8 tan θ = 2. If sin a = 35 and tan a < 0, the angle in standard position must be located in the 2nd quadrant. Sketch the angle in standard position and use Pythagorean Theorem to determine the length of the adjacent side. 96 (h)2 = (s1 )2 + (s2 )2 √ (3)2 = ( 5)2 + (s2 )2 √ 9 = 5 + (s2 )2 √ 4 = s2 2=s In the second quadrant, this value is negative. hyp adj 3 sec a = − 2 sec a = 3. To simplify cos4 x−sin4 x cos2 x−sin2 x , factor both the numerator and denominator using the diﬀerence of squares. cos4 x − sin4 x → factor cos2 x − sin2 x (cos2 x + sin2 x)(cos2 x − sin2 x) → factor (cos x + sin x)(cos x − sin x) (cos2 x + sin2 x)(cos x + sin x)(cos x − sin x) → simplify (cos x + sin x)(cos x − sin x) ( ( (( (( (sin (sin (− (+ (cos x) (cos x)( (cos2 x + sin2 x)( (x (x → simplify ( ( ( ( ( x)(cos x (( (sin (+ (cos (x (((− sin x) ( cos2 x + sin2 x → simplify → cos2 x + sin2 x = 1 cos4 x − sin4 x =1 cos2 x − sin2 x 1+sin x 4. To verify cos x sin x = sec x(csc x + 1), work with one side of the equation, making correct substitutions and performing accurate mathematical computations until both sides read the same. 1 + sin x = sec x(csc x + 1) cos x sin x 1 + sin x = sec x(csc x + 1) → working with LS. cos x sin x sin x 1 + = sec x(csc x + 1) → simplify cos sin x ( ) (x sin x) cos x 1 1 sin x + = sec x(csc x + 1) → simplify cos x sin x cos x sin x ( )( ) 1 1 1 + = sec x(csc x + 1) → reciprocal identities cos x sin x cos x sec x · csc x + sec x = sec x(csc x + 1) → common factor sec x(csc x + 1) = sec x(csc x + 1) 97 ( ) 5. To solve sec x + π2 + 2 = 0 for all values of x in the interval [0, 2π), the reciprocal identity for secant must be used. ( π) sec x + + 2 = 0 → solve 2 ) ( π + 2 − 2 = 0 − 2 → simplify sec x + 2 ( π) 1 sec x + = −2 → sec x = 2 cos x ( π) 1 cos x + =− 2 2 ( ) ( ( )) π 1 −1 −1 cos cos x + = cos − 2 2 π 2π 4π x+ = , → solve for x 2 3 3 π 2π x+ = 2 3 2π π π π − x+ − = 2 2 3 2 4π − 3π π x= = 6 6 ( ) 6. To solve 8 sin x2 − 8 = 0 for all values of x in the interval [0, 2π): π 4π = 2 3 π π 4π π x+ − = − 2 2 3 2 8π − 3π 5π x= = 6 6 x+ (x) 8 sin − 8 = 0 → solve (x) 2 8 sin − 8 + 8 = 0 + 8 → simplify 2 (x) 8 sin = 8 → simplify (2x ) 8 sin 2 8 = → simplify 8( ) 8 x sin = 1 → simplify ( ( x 2)) sin−1 sin = sin−1 (1) → simplify 2 π x = → solve 2 2 2x = 2π 2x 2π = 2 2 x=π 7. To solve 2 sin2 x + sin 2x = 0 for all values of x in the interval [0, 2π), will involve the double-angle identity for sine and the quotient identity for tangent. 2 sin2 x + sin 2x = 0 → double angle identity 2 sin2 x + 2 sin x cos x = 0 → common factor 2 sin x(sin x + cos x) = 0 → solve 98 Then2 sin x = 0 or sin x + cos x = 0 sin x + cos x = 0 → solve sin x + cos x − cos x = 0 → solve sin x = − cos x sin x cos x =− → quotient identity cos x cos x tan x = −1 2 sin x = 0 2 sin x 0 = 2 2 sin x = 0 sin−1 (sin x) = sin−1 (0) tan−1 (tan x) = tan−1 (−1) π x=− 4 x = 0, π The tangent function is negative in the 2nd and 4th quadrants. Therefore x = 3π 7π 4 , 4 7π All the values for x are: x = 0, 3π 4 , π, 4 8. To solve 3 tan2 2x = 1 for all values of x in the interval [0, 2π): 3 tan2 2x = 1 → solve 3 tan2 2x 1 = → simplify 3 3 1 3 tan2 2x = simplify 3 3 1 2 tan 2x = → simplify 3 √ 1 2 tan 2x = → both sides 3 √ √ 1 tan2 2x = → rationalize denominator 3 √ ( ) 1 3 simplify tan 2x = 3 3 √ 3 tan 2x = → solve 3 (√ ) 3 −1 −1 tan (tan 2x) = tan 3 The interval 0 ≤ x < 2π will be doubled since the equation deals with tan 2x. This will give the results in the interval 0 ≤ x < 4π 2x = π 7π 13π 19π , , , 6 6 6 6 To obtain the values of x, each of the above answers must be divided by 2. x= π 7π 13π 19π , , , 12 12 12 12 99 9. To determine the exact value of cos 157.5◦ , the half-angle formula for cosine must be used along with the angle 315◦ . √ θ cos θ + 1 cos = ± → θ = 315◦ 2 2 √ 315◦ cos 315◦ + 1 cos =± → 157.5◦ (2nd quadrant(-)) 2 2 √√ 2 315◦ 2 +1 =± → common denominator cos 2 2 √√ (2) 2 315◦ 2 + 2 1 cos =± → simplify 2 2 √√ 2+2 315◦ 2 cos =± → simplify 2 2 v( )( ) u √ u 315◦ 2+2 1 t cos =± → simplify 2 2 2 v( ) u √ u 315◦ 2+2 t cos =± → simplify 2 4 √√ 315◦ 2+2 cos =± 2 2 10. To determine the exact value of 13π 12 , the sine formula for the sum of angles must be used. The angle 10π 13π 13π can be expressed as the sum of 12 12 and 12 . 10π 12 3π →b= ( ) ( ) ( ) ( 12 ) ( ) 10π 3π 10π 3π 10π 3π sin + = sin cos + cos sin → simplify 12 12 12 12 12 12 ( ) ( ) (π ) (π) (π) 5π π 5π sin + = sin cos + cos sin → simplify 6 4 6 4 6 4 ) ( ) (√ ) ( √ ) (√ ) ( 1 5π π 2 3 2 + = + − → simplify sin 6 4 2 2 2 2 ( ) (√ ) ( √ ) 5π π 2 6 sin + = + − → simplify 6 4 4 4 √ ( ) √ 5π π 4− 6 sin + = 6 4 4 sin(a + b) = sin a cos b + cos a sin b → a = 11. To write 4(cos 5x + cos 9x) as a product, the sum to product formula for cosine will be used. 100 ( cos α + cos β = 2 cos ( cos 5x + cos 9x = 2 cos α+β 2 ) 5x + 9x 2 ( · cos α−β 2 ) ( · cos ) → α = 5x → β = 9x ) 5x − 9x → simplify 2 cos 5x + cos 9x = 2 cos(7x) · cos(−2x) 12 To simplify cos(x − y) cos y − sin(x − y) sin y, the diﬀerence formulas for both cosine and sine must be applied. In addition the Pythagorean Identity sin2 x + cos2 x = 1 will be used. cos(x − y) = cos x cos y + sin x sin y sin(x − y) = sin x cos y − cos x sin y (cos x cos y + sin x sin y) cos y − (sin x cos y − cos x sin y) sin y → simplify cos x cos2 y + sin x sin y cos y − sin x sin y cos y − cos x sin2 y → simplify cos x cos2 y + cos x sin2 y → common factor (cos x) cos x(cos2 y + sin2 y) → sin2 x + cos2 x = 1 cos x(1) ∴ cos(x − y) cos y − sin(x − y) sin y = cos x 13. To simplify the trigonometric expression sin the sum formula for cosine will both be used. ( 4π 3 ) ( − x + cos x + 101 5π 6 ) the diﬀerence formula for sine and 4π 3 →b=x sin(a − b) = sin a cos b − cos a sin b → a = ( sin 4π −x 3 ) = sin 4π 4π cos x − cos sin x 3 3 cos(a + b) = cos a cos b − sin a sin b → a = x 5π →b= 6 ( ) 5π 5π 5π cos x + − sin x sin = cos x cos 6 6 6 4π 4π 5π 5π sin cos x − cos sin x + cos x cos − sin x sin → simplify 3 ) 3 6 ( √ (6 √ ) ( ( ) ) 3 1 3 1 − cos x − − sin x + cos x − − sin x → simplify 2 2 2 2 ( √ ) √ 1 1 3 3 cos x + sin x − cos x − sin x → simplify − 2 2 2 2 (√ ) 3 −2 cos x → simplify 2 (√ ) √ 3 −2 cos x = − 3 cos x 2 ( ) ( ) √ 4π 5π sin − x + cos x + = − 3 cos x 3 6 14. To derive a formula for sin 6x, the function must be expressed as sin (4x + 2x). This means that the sum formula for sine must be used as well as the double angle formula for sine and cosine. sin(a + b) = sin a cos b + cos a sin b → a = 4x → b = 2x sin(4x + 2x) = sin 4x cos 2x + cos 4x sin 2x → expand sin(4x + 2x) = sin(2x + 2x) cos 2x + cos(2x + 2x) sin 2x → expand sin(4x + 2x) = cos 2x(sin 2x cos 2x + cos 2x sin 2x) + sin 2x(cos 2x cos 2x − sin 2x sin 2x) → expand sin(4x + 2x) = sin 2x cos2 x + cos2 2x sin 2x + sin 2x cos2 2x − sin3 x → simplify sin(4x + 2x) = 3 sin 2x cos2 x − sin3 x → common factor sin(4x + 2x) = sin 2x(3 cos2 x − sin2 x) → double angle formula [ ] sin(4x + 2x) = 2 sin x cos x 3(cos2 x − sin2 x)2 − (2 sin x cos x)2 → simplify [ ] sin(4x + 2x) = 2 sin x cos x 3(cos4 x − 2 cos2 x sin2 x + sin4 x) − 4 sin2 x cos2 x → simplify [ ] sin(4x + 2x) = 2 sin x cos x 3 cos4 x − 6 cos2 x sin2 x + 3 sin4 x − 4 sin2 x cos2 x → simplify [ ] sin(4x + 2x) = 2 sin x cos x 3 cos4 x + 3 sin4 x − 10 sin2 x cos2 x → simplify sin(4x + 2x) = 6 sin x cos5 x + 6 sin5 x cos x − 20 sin3 x cos3 x 102 Chapter 4 TE Inverse Functions and Trigonometric Equations - Solution Key 4.1 Inverse Functions and Trigonometric Equations General Deﬁnitions of Inverse Trigonometric Functions Review Exercises 1. a) This graph represents a one-to-one function because a vertical line would cross the graph at only one point and a horizontal line would also cross the graph at only one point. Therefore the graph passes both the vertical line test and the horizontal line test. At this point students do know whether or not the function has an inverse that is a function. As a result, it is ﬁne to accept whatever answer the students present as long as they justify their answer. 103 b) This graph represents a function because it passes the vertical line test. However, the graph does not pass the horizontal line test. It does not have an inverse that is a function. c) The above graph passes the horizontal line test only. It fails the vertical line test. Therefore, this graph does not represent a one-to-one function. It does however, have an inverse that is a function. 2. To calculate the measure of the angle that the ladder makes with the ﬂoor, the trigonometric ratio for cosine must be used. The ladder is the hypotenuse of the right triangle and the distance from the wall is the adjacent side with respect to the reference angle. adj hyp 4 cos θ = 9 cos θ = 0.4444 cos θ = cos−1 (cos−1 θ) = cos−1 = (0.4444) θ ≈ 63.6◦ 104 ( ) 1. sin−1 π2 does not exist. If π is considered as having an approximate value of 3.14, then domain of the sine function is [−1, 1]. 3.14 2 ≈ 1.57. The 2. tan−1 (−1) does exist. The graph of tan−1 (−1) can be done on the graphing calculator. The exact value is − π4 . y = tan−1 3. cos−1 π 3. (1) y = cos−1 2 does exist. The graph of cos−1 (1) 2 can be done on the graphing calculator. The exact value is (1) 2 Ranges of Inverse Circular Functions Review Exercises To determine the exact values of the following functions, the special triangles may be used or the unit circle. The special triangles may be easier for students to sketch and the answers can be readily converted to radians or degrees if necessary. 1. a) cos 120◦ An angle of 120◦ has a related angle of 60◦ in the 2nd quadrant. The cosine function is negative adj in this quadrant. Using the special triangle, the exact value of cos 120◦ is hyp = − 12 105 3π π ◦ ◦ nd b) csc 3π quadrant. Cosecant is the 4 . An angle of 4 rad (135 ) has a related angle of 4 rad (45 ) in the 2 nd reciprocal of the sine function and is positive in the 2 quadrant. Therefore, using the special triangle, if √ √1 then csc 3π = = sin 3π 2. 4 4 2 5π π ◦ ◦ th c) tan 5π quadrant. The tangent 3 . An angle of 3 rad (300 ) has a related angle of 3 rad (60 ) in the 4 th function has a negative value in the 4 quadrant. Using the special triangle, the exact value of tan 5π 3 is √ opp 3 adj = − 1 . 106 a) ◦ 90 π Using this diagram shows that cos−1 (0) = 90◦ or 180 ◦ = 2 rad √ th b) tan1 (− 3) = −60◦ in either the 2nd quadrant since the tangent function is negative √ or the 41 quadrant √ 1 in these quadrants. The exact value of tan (− 3) is tan (− 3) = −60◦ or − π3 rad ( ) c) sin−1 − 12 = −30◦ in either the 3rd or the 4th quadrant since the sine function is negative in these ( ) quadrants. The exact value of sin−1 − 12 = −30◦ or − π6 rad is Review Exercises 1. The graphs of y = x6 + 2x2 − 8 and y = x can be graphed using the TI-83. From the graph, it is obvious that the graph of y = x6 + 2x2 − 8 would not reﬂect across the line y = x as a mirror image. Therefore the function is not invertible. 107 b) The graphs of y = cos(x3 ) and y = x are shown below as displayed on the TI-83. The graph of the inverse x = cos(y 3 ) is shown below as it appears when added to the above graph on the TI-83. The function y = cos(x3 ) is invertible because its inverse, x = cos(y 3 ), is the mirror image of y = cos(x3 ) reﬂected across the line y = x. 2. To prove that the functions f (x) = 1 − f (f −1 (x)) = x. and f −1 (f (x)) = x. 1 x−1 and f −1 (x) = 1 + 108 1 1−x are inverses, prove algebraically that f (f −1 (x)) = 1 − ( 1 1 1−x 1+ ) 1 ) f (f −1 (x)) = 1 − ( ( 1−x 1 1−x + f (f −1 (x)) = 1 − f (f −1 (x)) = 1 − f (f −1 (x)) = 1 − f (f −1 (x)) = 1 − f (f −1 → common denominator −1 1 1−x ) −1 → simplify 1 → simplify −1 1 ( ) → simplify → common denominator − 1−x 1−x 1 1−x+1 1−x 2−x 1−x 1 2−x−1+x 1−x 1 → simplify → simplify 1 1−x [ ( )] 1−x (x)) = 1 − 1 → simplify 1 f (f −1 (x)) = 1 − (1 − x) → simplify f (f −1 (x)) = 1 − 1 + x → simplify f (f −1 (x)) = x f −1 (f (x)) = 1 + f −1 (f (x)) = 1 + f −1 (f (x)) = 1 + 1 ( 1− 1− 1 x−1 ) → common denominator 1 ( ( ) 1 − 1 x−1 x−1 − 1− f −1 (f (x)) = 1 + ( 1 f −1 (f (x)) = 1 + ( ( 1 x−1−1 x−1 x−1 x−1 1 1 x−1 ) 1 − 1 x−1 ) → simplify ) → simplify ( x−2 x−1 ) → simplify → common denominator ) → simplify [ ( )] x−1 f −1 (f (x)) = 1 + 1 → simplify 1 f −1 (f (x)) = 1 + x − 1 → simplify f −1 (f (x)) = x Derive Properties of Other Five Inverse Circular Functions in Terms of Arctan Review Exercises 1. 109 a) Using this triangle will determine a value for tan−1 (x). opp adj x tan θ = 1 (tan θ) = tan−1 (x) tan θ = tan−1 θ = tan−1 (x) cos2 (tan−1 x) = cos2 (θ) Using the same triangle, determine the length of the hypotenuse. (h)2 = (s1 )2 + (s2 )2 (h)2 = (x)2 + (1)2 √ adj hyp 1 cos θ = √ 2 x +1 ( )2 1 cos2 θ = √ x2 + 1 1 cos2 θ = 2 x +1 1 2 −1 ∴ cos (tan x) = 2 x +1 cos θ = b) cot(tan−1 x2 ) − cot2 (tan−1 x) As shown above, tan−1 x = θ 110 (h)2 = x2 + 1 √ (h)2 = x2 + 1 √ h = x2 + 1 adj 1 = hyp x ( )2 1 1 2 = 2 cot θ = x x 1 ∴ cot(tan−1 x2 ) = 2 x cot θ = 2. The graph of can be displayed using the TI-83. The domain is the set of all real numbers except π 2 ( ) + kπ where k is an integer and the range is − π2 , π2 Review Exercises ) (( π ) 1. To prove sin 2 − θ = cos θ the cofunction identities for sine and cosine must be used. ) − θ = cos θ → cofunction identities 2 (π ) cos − θ = sin θ (( π 2) ) (π (π )) sin − θ = cos − −θ → simplify 2) 2 2 ) (( π ) (π π sin − θ = cos − + θ → simplify 2 2 2 (( π ) ) sin − θ = cos(0 + θ) 2) (( π ) sin − θ = cos(θ) 2 sin 2. If sin (π 2 (π ) ( ) − θ = 0.68 and sin π2 − θ = cos(θ) then − sin (π ) − θ = cos(−θ) 2 ∴ cos(−θ) = −0.68 Review Exercises 1. To determine the exact values of the following inverse functions, the special triangles can be used. 111 (√ ) 3 a) cos−1 From the triangles, it can be veriﬁed that cos θ(30◦ ) = 2 (√ ) 3 cos−1 is π6 . 2 adj hyp √ = 3 2 . The exact value of √ √ b) sec−1 ( 2). The secant function is the reciprocal of the cosine function. Therefore, sec θ(45◦ ) = hyp = 12 . adj √ The exact value of sec−1 ( 2) is π4 . √ c) sec−1 (− 2). The secant function is the reciprocal of the cosine function and is therefore negative in the 2nd and 3rd quadrants. An angle of 45◦ in standard position in the 2nd quadrant is an angle of 225◦ . √ √ hyp 2 ◦ sec θ(225 ) = adj = − 1 The exact value of sec−1 ( 2) is 5π 4 . Review Exercises ( −1 ( 5 )) 1. To evaluate sin cos in the 1st quadrant. Working backwards, the previous 13 , the angle(is located ( ) ) 5 −1 5 −1 −1 5 line to cos (cos θ) = cos 13 is cos 13 . Thus, cos θ = 13 . ( sin cos−1 ( 5 13 )) = sin θ sin θ = This solution can be veriﬁed using technology: 112 12 . 13 Revisiting Revisiting y = c + a cos b(x − d) Review Exercises 1. The transformations of y = cos x are the vertical reﬂection; vertical stretch; vertical translation; horizontal stretch and horizontal translation. These changes can be used to write the equation to model a graph of a sinusoidal curve. The simplest way to present these transformations is show them in a list. V.R. = No V.S. = 5 − −1 =3 2 V.T. = 2 H.S. = 210◦ − 30◦ 1 = 360◦ 2 H.T. = 30◦ The equation that would model the graph of y = cos x that has undergone these transformations is y = 3 cos(2(x − 30◦ )) + 2 Review Exercises 1. This problem is an example of an application of solving the equation y = c + a cos b(x − d) in terms of x. The problem that is presented should be sketched as a graph to facilitate obtaining an equation to model the curve. Once this has been done, the equation can then be entered into the TI-83 and the trace function can be used to estimate a value for x. The following graph was done on the calculator and it shows an estimate of 3.34 seconds for x. y = 32 + cos 6.28 8 6.28 y = 32 + cos 8 ( ) 12 x− → equation 6.28 ( ) 12 x− → solve for x 6.28 113 [ ] cos−1 y−c 6.28 12 a x= + d → y = 40, c = 32, b = ,d = b 8 6.28 x= x= cos−1 [ cos−1 8 20 [ 40−32 ] 20 6.28 8 ] 6.28 8 + + 12 simplify 6.28 12 → simplify u sin g T I − 83 6.28 x ≈ 3.39 seconds Solving Trigonometric Equations Analytically Review Exercises 1. To solve the equation sin 2θ = 0.6 for 0 ≤ θ < 2π, involves determining all the possible values for sin 2θ = 0.6 for 0 ≤ θ < 4π and then dividing these values by 2 to obtain the values for π. The angle is measured in radians since the domain is given in these units. sin 2θ = 0.6 → determine reference angle. α = sin−1 (0.6) α = 0.6435 The angles for 2θ will be in quadrants 1, 2, 5, 6. 2θ = 0.6435, π − 0.6435, 2π + 0.6435, 3π − 0.6435 2θ = 0.6435, 2.4980, 6.9266, 8.7812 The angles for θ in the domain [0, 2π) are: θ = 0.3218, 1.2490, 3.4633, 4.3906 It is not necessary, but these results can be conﬁrmed by using the TI-83 calculator to graph the function. 1 2. To solve the equation cos2 x = 16 over the interval [0, 2π) involves applying the fact that the square root of a number can be positive or negative. This will allow the equation to be solved for all possible values. 114 √ 1 → Both sides 16 √ √ 1 cos2 x = → simplify 16 1 cos x = ± 4 ( ) 1 cos−1 (cos x) = cos−1 4 cos2 x = Then x = 1.3181radians → 1st eqadrant x = 2π − 1.3181 x = 4.9651radians → 4th eqadrant ( ) 1 −1 −1 − cos (cos x) = cos 4 Or x = 1.8235radians → 1st eqadrant x = 2π − 1.8235 → 3rd eqadrant x = 4.4597radians Once again, the results can be conﬁrmed by graphing the function using the TI-83. 3. To solve the equation sin 4θ − cos 2θ = 0 for all values of θ such that 0 ≤ θ ≤ 2π involves using the double angle identity for sine. sin 4θ − cos 2θ = 0 2 sin 2θ cos 2θ − cos 2θ = 0 → common factor cos 2θ(2 sin 2θ − 1) = 0 → simplify Then cos 2θ = 0over the interval [0, 4π] π 3π 5π 7π , , , , → ÷2 2 2 2 2 π 3π 5π 7π θ= , , , 4 4 4 4 2θ = Or 115 2 sin 2θ − 1 = 0 2 sin 2θ = 1 1 sin 2θ = → over the interval [0, 4π] 2 π 5π 13π 17π 2θ = , , , → ÷2 6 6 6 6 π 5π 13π 17π θ , , , 12 12 12 12 Once again, the results can be conﬁrmed by graphing the function using the TI-83. 4. To solve the equation tan 2x − cot 2x = 0 over the interval 0◦ ≤ x < 360◦ will involve using the reciprocal identity for cotangent and applying the fact that the square root of a number can be positive or negative. This will allow the equation to be solved for all possible values. tan 2x − cot 2x = 0 tan 2x − cot 2x = 0 → cot x = tan 2x − 1 tan x 1 = 0 → simplify tan 2x 1 = (tan 2x)0 → simplify tan 2x 1 tan 2x(tan 2x) − (tan 2x) = (tan 2x)0 → simplify tan 2x tan2 2x − 1 = 0 → simplify √ tan2 2x = 1 → Both sides √ √ tan2 2x = 1 tan 2x = ±1 tan 2x(tan 2x) − (tan 2x) Then tan 2x = 1 over the interval [0◦ , 720◦ ). The tangent function is positive in the 1st , 3rd , 5th and 7th quadrants. 2x = 45◦ , 225◦ , 405◦ , 5825◦ → ÷2 x = 22.5◦ , 112.5◦ , 202.5◦ , 292.5◦ Or tan 2x = −1 over the interval [0◦ , 720◦ ). The tangent function is negative in the 2nd , 4th , 6th, and 8th quadrants. 2x = 135◦ , 315◦ , 495◦ , 675◦ x = 67.5◦ , 157.5◦ , 247.5◦ , 337.5◦ 116 Once again, the results can be conﬁrmed by graphing the function using the TI-83. Review Exercises 1. To solve sin2 x − 2 sin x − 3 = 0 for the values of x that are within the domain of the sine function, involves factoring the quadratic equation and determining the values that fall within the domain of [0, 2π] or [0, 360◦ ]. sin2 x − 2 sin x − 3 = 0 sin2 x − 2 sin x − 3 = 0 → factor (sin x + 1)(sin x − 3) = 0 → simplify Then sin x + 1 = 0 sin x = −1 sin−1 (sin x) = sin−1 (−1) 3π x = 270◦ or 2 Or sin x − 3 = 0 sin x = 3 Does not exist. It is not in the range [−1, 1] of the sine function. 2. To solve the equation tan2 x = 3 tan x for the principal values of x involves factoring the quadratic equation and determining the values that fall within the domain of the function. tan2 x = 3 tan x tan2 x − 3 tan x = 0 → common factor tan x(tan x − 3) = 0 → simplify Then Or tan x − 3 = 0 tan x = 0 tan x(tan x) = tan −1 tan x = 3 (0) ◦ tan−1 (tan x) = tan−1 (3) x = 71.5◦ x=0 3. To solve the equation sin x = cos x2 over the interval [0◦ , 360◦ ) requires the use of the Pythagorean Identity sin2 θ + cos2 θ = 1 and the half-angle identity for cosine. 117 x 2 √ x cos x + 1 sin x = cos → ± 2 2 √ cos x + 1 sin x = ± → squre both sides 2 ( √ )2 cos x + 1 2 → squre both sides (sin x) = ± 2 sin x = cos cos x + 1 → sin2 x + cos2 x = 1 2 cos x + 1 1 − cos2 x = → simplify (2 ) cos x + 1 2 2(1 − cos x) = 2 → simplify 2 ( ) cos x + 1 2(1 − cos2 x) = 2 → simplify 2 sin2 x = 2 − 2 cos2 x = cos x + 1 → simplify 2 − 2 cos2 x − cos x − 1 = 0 → simplify −2 cos2 x − cos x + 1 = 0 → ÷(−1) 2 cos2 x + cos x − 1 = 0 → factor (2 cos x − 1)(cos x + 1) = 0 Then Or 2 cos x − 1 = 0 1 cos x = 2 cos x + 1 = 0 cos x = −1 ( ) 1 cos−1 (cos x) = cos−1 2 cos−1 (cos x) = cos−1 (−1) Cosine is positive in the 1st and 4th quadrants. Cosine is negative in the 2nd and 3rd quadrant. x = 60◦ , 300◦ x = 180◦ 4. To solve the equation 3 − 3 sin2 x = 8 sin x over the interval [0, 2π] requires factoring the quadratic equation and solving for all the solutions. 3 − 3 sin2 x = 8 sin x 3 − 3 sin2 x − 8 sin x = 8 sin x − 8 sin x → simplify −3 sin2 x − 8 sin x + 3 = 0 → ÷(−1) 3 sin2 x + 8 sin x − 3 = 0 → factor (3 sin x − 1)(sin x + 3) = 0 118 Then Or 3 sin x − 1 = 0 1 sin x = 3 sin x + 3 = 0 sin x = −3 ( ) 1 −1 −1 sin (sin x) = sin 3 sin−1 (sin x) sin−1 (−3) Sine is positive in the 1st and 2nd quadrants. Does not exist. x = 0.3398radians x = π − 0.3398 x = 2.8018radians Review Exercises 1. To solve the equation 2 sin x tan x = tan x + sec x for all values of xε[0, 2π] requires the use of the quotient identity for tangent and the reciprocal identity for secant. 2 sin x tan x = tan x + sec x 1 sin x ; sec x = 2 sin x tan x = tan x + sec x → tan x = cosx cos x ( ) ( ) ( ) sin x sin x 1 2 sin x = + → simplify cosx cosx cosx sin2 x sin x + 1 2 = → simplify cosx cosx ( 2 ) ( ) sin x sin x + 1 2 (cos x) = (cos x) → simplify cos x cos x ( ) ( 2 ) sin x = sin x + 1 → simplify x) x) (cos (cos 2 cos x cos x 2 sin2 x = sin x + 1 → simplify 2 sin2 x − sin x − 1 = 0 → factor (2 sin x + 1)(sin x − 1) = 0 Then 2 sin x + 1 = 0 1 sin x = − 2 Or sin x − 1 = 0 ( sin−1 (sin x) = sin−1 − 1 2 sin x = 1 ) sin−1 (sin x) sin−1 (1) Sine is negative in the 3rd and 4th quadrants. x= x= π radians 2 7π 11π and radians 6 6 2. To solve the equation cos 2x = −1 + cos2 x for all values of x can be simply solved by using the double angle formula for cosine. 119 cos 2x = −1 + cos2 x cos 2x = −1 + cos2 x → cos(2x) = 2 cos2 x − 1 2 cos2 x − 1 = −1 + cos2 x → simplify 2 cos2 x − 1 + 1 − cos2 x = 0 → simplify √ cos2 x = 0 → Both sides √ √ cos2 x = 0 cos x = 0 cos−1 (cos x) = cos−1 (0) x= π 2 and for all values of x, x = π 2 + kπ, where kεI. 3. To solve the equation 2 cos2 x + 3 sin x − 3 = 0 for all values of x over the interval [0, 2π] requires the use of the Pythagorean Identity sin2 θ + cos2 θ = 1. 2 cos2 x + 3 sin x − 3 = 0 2 cos2 x + 3 sin x − 3 = 0 → sin2 x + cos2 x = 1 2(1 − sin2 x) + 3 sin x − 3 = 0 → expand 2 − 2 sin2 x + 3 sin x − 3 = 0 → simlify −2 sin2 x + 3 sin x − 1 = 0 → ÷(−1) 2 sin2 x − 3 sin x + 1 = 0 → factor (2 sin x − 1)(sin x − 1) = 0 Then Or 2 sin x − 1 = 0 1 sin x = 2 sin x − 1 = 0 sin−1 (sin x) = sin−1 sin x = 1 1 2 sin−1 (sin x) = sin−1 (1) Sine is positive in the 1st and 2nd quadrants x= x= π 2 5π π and radians 6 6 Review Exercises 1. To solve the equation 3 cos2 x − 5 sin x = 4 for all values of x over the interval 0◦ ≤ x ≤ 360◦ will require writing the equation in terms of sine by using the Pythagorean Identity sin2 θ + cos2 θ = 1 and then using the quadratic formula to solve the equation. 120 3 cos2 x − 5 sin x = 4 3 cos2 x − 5 sin x = 4 → sin2 x + cos2 x = 1 3(1 − sin2 x) − 5 sin x = 4 → expand 3 − 3 sin2 x − 5 sin x = 4 → simplify 3 − 3 sin2 x − 5 sin x − 4 = 4 − 4 → simplify −3 sin2 x − 5 sin x − 1 = 0 → ÷(−1) 3 sin2 x + 5 sin x + 1 = 0 → ÷(−1) Let y = sin x 3y 2 + 5y + 1 = 0 a=3b=5c=1 √ −b ± b2 − 4ac y= 2a √ −5 ± (5)2 − 4(3)(1) y= → simplify 2(3) √ −5 ± 13 y= → simplify 6 Then Or −5 + 13 y= 6 y ≈ −0.2324 √ −5 − 13 y= 6 y ≈ −1.4342 y = sin x sin x = −0.2324 y = sin x sin x = −1.4342 sin−1 (sin x) = sin−1 (−0.2324) sin−1 (sin x) = sin−1 (−1.4342) Sine is negative in the 3rd and 4th quadrants x ≈ 193.5◦ and x ≈ 346.5◦ Does not exist. √ 2 2. The quadratic formula [ πmust ] be used to solve the trigonometric equation tan x + tan x + 2 = 0 for values π of x over the interval − 2 , 2 tan2 x + tan x + 2 = 0 tan2 x + tan x + 2 = 0 Let y = tan x y2 + y + 2 = 0 a=1b=1c=2 √ −b ± b2 − 4ac Y = 2a √ −1 ± (1)2 − 4(1)(−2) Y = → simplify 2(1) √ −1 ± 9 → simplify Y = 2(1) 121 Then Or −1 + 9 y= → simplify 2(1) y=1 y = tan x √ −1 − 9 y= → simplify 2(1) y = −2 y = tan x tan x = 1 tan x = −2 tan −1 x= √ −1 (tan x) = tan tan−1 (tan x) = tan−1 (−2) (1) π + kπ 4 x = arctan(−2) + kπ 3. To solve the equation 5 cos2 θ − 6 sin θ = 0 over the interval [0, 2π] involves using the Pythagorean Identity sin2 θ + cos2 θ = 1 and the quadratic formula. 5 cos2 θ − 6 sin θ = 0 5 cos2 θ − 6 sin θ = 0 → sin2 θ + cos2 θ = 1 5(1 − sin2 θ) − 6 sin θ = 0 → expand 5 − 5 sin2 θ − 6 sin θ = 0 → simplify −5 sin2 θ − 6 sin θ + 5 = 0 → ÷(−1) 5 sin2 θ + 6 sin θ − 5 = 0 → solve Let y = sin x 5y 2 + 6y − 5 = 0 a = 5 b = 6 c = −5 √ −b ± b2 − 4ac y= 2a √ −6 ± (6)2 − 4(5)(−5) y= → simplify 2(5) √ −6 ± 136 y= → simplify 10 Then −6 + 11.66 y= → simplify 10 y = 0.566 Or y= y = sin x sin x = 0.566 y = sin x sin x = −1.766 sin−1 (sin x) = sin−1 (0.566) x ≈ 0.6016radians ± 2π sin−1 (sin x) = sin−1 (1.766) Dose not exit −6 − 11.66 → simplify 10 y = −1.766 x ≈ 2.5399radians ± 2π Solve Equations (with double angles) Review Exercises 122 1. If tan x = 34 and 0◦ < x < 90◦ , the angle is in standard position in the 1st quadrant. The triangle is a 3 − 4 − 5 triangle which makes sin x = 53 and cos x = 54 . The value of tan 2x can be found by using the double angle formula for tangent. a) 2 tan x 1 − tan2 x ( ) 2 34 tan (2x) = ( )2 → simplify 1 − 34 ( ) 3 2 42 tan (2x) = 9 → common deno min ator 1 − 16 tan (2x) = 3 tan (2x) = tan (2x) = tan (2x) = 1 ( 16 )2 16 − 9 16 → simplify 3 2 16−9 → simplify 16 3 2 7 → simplify 16 ( ) 3 16 tan (2x) = · simplify 2 7 24 ≈ 3.4286 tan (2x) = 7 b) The value of sin 2x can be found by using the double angle formula for sine and the values of sin x which is 53 and of cos x which is 45 . sin (2x) = 2 sin x cos x 3 5 4 → cos x = 5 ( ) ( ) 3 4 sin (2x) = 2 · → simplify 5 5 24 sin (2x) = = 0.960 25 sin (2x) = 2 sin x cos x → sin x = c) The value of cos 2x can be found by using the double angle formula for cosine and the values of sin x which is 53 and of which is 45 . 123 cos (2x) = cos2 x sin2 x 3 5 4 → cos x = 5 cos (2x) = cos2 x − sin2 x → sin x = ( )2 ( )2 4 3 cos (2x) = · → simplify 5 5 16 − 9 → simplify cos (2x) = 25 7 cos (2x) = = 0.280 25 2. To prove that 2 csc(2x) = csc2 x tan x is an identity, work with the left side. The reciprocal identity for sine must be used as well as the double angle formula for sine. 2 csc(2x) = csc2 x tan x 2 csc(2x) = csc2 ( ) 1 2 = csc2 sin(2x) 2 = csc2 sin(2x) 2 = csc2 2 sin x cos x 2 = csc2 2 sin x cos x 1 = csc2 sin x cos x ( ) 1 sin x = csc2 sin x sin x cos x sin x = csc2 2 sin x cos x ( ) ( ) sin x 1 · = csc2 cos x sin2 x x tan x → csc x = 1 sin x x tan x → simplify x tan x → double angle formula x tan x → simplify x tan x → simplify x tan x → multiply left side by sin x sin x x tan x → simplify x tan x → express as factors 1 = cos x sin x sin x = tan x → cos x x tan x → csc2 x tan x = csc2 x tan x b) To prove that cos4 θ − sin4 θ = cos 2θ is an identity, work with the left side. The left side must be factored by using the diﬀerence of two squares and then the Pythagorean Identity sin2 θ + cos2 θ = 1 must be applied. 124 cos4 θ − sin4 θ = cos 2θ cos4 θ − sin4 θ = cos 2θ → factor (cos2 θ − sin2 θ)(cos2 θ − sin2 θ) = cos 2θ → sin2 θ + sin2 θ = 1 1(cos2 θ − sin2 θ) = cos 2θ → simplify (cos2 θ − sin2 θ) = cos 2θ → cos2 θ − sin2 θ = cos 2θ cos 2θ = cos 2θ sin 2x c) To prove that 1+cos 2x = tan x is an identity, work with the left side. The double angle formula for sine and the double angle formula for cosine must be used. sin 2x = tan x 1 + cos 2x sin 2x = tan x → sin 2x = 2 sin x cos x 1 + cos 2x → cos 2x = 1 − 2 sin2 x 2 sin x cos x = tan x → simplify 1 + (1 − 2 sin2 x) 2 sin x cos x = tan x → common factor 2 − 2 sin2 x 2 sin x cos x = tan x → sin2 x + cos2 x = 1 2(1 − sin2 x) 2 sin x cos x = tan x → factor 2 cos2 x 2 sin x cos x = tan x → simplify 2(cos x)(cos x) 2 cos x sin x sin x = tan x → = tan x cos x 2(cos x) (cos x) sin x = tan x cos x tan x = tan x 3. To solve the trigonometric equation cos 2θ = 1 − 2 sin2 θ such that −π ≤ θ < π involves using the double angle formula for cosine. 4. To solve the trigonometric equation cos 2x = cos x such that 0 ≤ x < π involves using the double angle formula for cosine. cos 2x = cos x cos 2x = cos x → 2 cos2 x − 1 = cos 2x 2 cos2 x − 1 = cos x → simplify 2 cos2 x − 1 − cos x = 0 → simplify 2 cos2 x − cos x − 1 = 0 → factor (2 cos x + 1)(cos x − 1) = 0 125 Then 2 cos x + 1 = 0 1 cos x = − 2 Or cos x − 1 = 0 cos x = 1 ( ) 1 −1 −1 cos (cos x) = cos − 2 cos−1 (cos x) = cos−1 (1) Cosine is negative in the 2nd quadrant 2π radians x= 3 =0 Review Exercises 1. a) To determine the exact value of sin 67.5◦ , the half-angle identity for sine will be used with an angle of 135◦ 2 . The special triangles will also be used. 126 v ( ) u √ u 1 √2 + t 135◦ 2 =± sin 2 2 √ θ 1 − cos θ sin = ± 2 2 √ θ 1 − cos θ sin = ± → θ = 135◦ 2 2 √ 135◦ 1 − cos 135◦ 1 sin =± → cos 135◦ = − √ 2 2 2 v ) ( u u 1 − − √1 t 135◦ 2 =± → simplify sin 2 2 √ 1 + 12 135◦ sin =± → common deno min ator 2 2 √1 2 → simplify √√ 2+1 √ 135◦ 2 sin =± → simplify 2 2 v( ) ( ) u √ u 135◦ 2 + 1 1 √ sin = ±t · → simplify 2 2 2 v( ) u √ ◦ u 135 2 + 1 √ sin = ±t → rationalize deno min ator 2 2 2 v( ) (√ ) u √ ◦ u 135 2 + 1 2 √ √ → simplify sin = ±t 2 2 2 2 v( u √ √ ) ◦ u 135 4 + 2 √ sin → simplify = ±t 2 2 4 v( u √ ) ◦ u 2+ 2 135 sin → simplify = ±t 2 4 √√ 135◦ 2+2 sin =± 2 2 An angle of 67.5◦ is located in the 1st quadrant and the sine of an angle in this quadrant is positive. 135◦ = ∴ sin 2 √√ 2+2 2 b) To determine the exact value of tan 165◦ , the half-angle identity for tangent will be used with an angle ◦ of 330 2 . The special triangles will also be used. 127 θ 1 − cos θ = 2 sin θ θ 1 − cos θ tan = → θ = 330◦ 2 sin θ √ 1 − cos 330◦ 330◦ 3 ◦ tan = → cos 330 = 2 sin 330◦ 3 1 → sin 330◦ = − 2 √ 3 ◦ 1− 2 330 tan = → commom denominator 2 − 12 ( ) √ 1 22 − 23 330◦ = → simplify tan 2 − 12 tan √ 2− 3 330◦ tan = 21 → simplify 2 −2 √ ( ) ◦ 330 2− 3 2 tan = · − → simplify 2 2 1 √ 330◦ −4 + 2 3 tan = → simplify 2 2 √ 330◦ −42 + 2 3 tan = 2 2 √ 330◦ tan = −2 + 3 2 ( ) ( ) 2. To prove that sin x tan x2 + 2 cos x = 2 cos2 x2 work with both sides of the equation and use the half-angle identity for cosine and the half-angle identity for tangent. sin x tan Left Side: sin x tan (x) 2 + 2 cos x = 2 cos2 (x) + 2 cos x 2 (x) ( x ) 1 − cos x sin x tan + 2 cos x → tan = 2 2 sin x ( ) 1 − cos x sin x + 2 cos x → simplify sin x ( ) 1 − cos x sin x + 2 cos x → simplify sin x 1 − cos x + 2 cos x → simplify 1 + cos x (x) 2 Right Side (x) 2 cos2 2 √ (x) x cos x + 1 2 2 cos → cos = ± 2 2 2 ( √ )2 cos x + 1 2 ± → simplify 2 ( ) cos x + 1 2 → simplify 2 ( ) cos x + 1 2 2 cos x + 1 Since both sides of the equation equal 1 + cos x, they are equal to each other. 128 3. To solve the trigonometric equation cos x2 = 1 + cos x such that 0 ≤ x < 2π the half- angle identity for cosine must be applied. x 2 x cos 2 √ cos x + 1 ± 2 ( √ )2 cos x + 1 ± 2 cos = 1 + cos x √ x cos x + 1 = 1 + cos x → cos = ± 2 2 = 1 + cos x → square both sides = (1 + cos x)2 → expand cos x + 1 = 1 + 2 cos x + cos2 x → simplify 2 ) ( cos x + 1 2 = 2(1 + 2 cos x + cos2 x) → simplify 2 ) ( cos x + 1 = 2(1 + 2 cos x + cos2 x) → simplify 2 2 cos x + 1 = 2 + 4 cos x + 2 cos2 x → simplify cos x − cos x + 1 − 1 = 2 + 4 cos x + 2 cos2 x − cos x − 1 → simplify 0 = 2 cos2 x + 3 cos x + 1 → simplify 2 cos2 x + 3 cos x + 1 = 0 → solve (2 cos x + 1)(cos x + 1) = 0 Then Or 2 cos x + 1 = 0 1 cos x = − 2 cos x + 1 = 0 cos x = −1 ( ) 1 −1 −1 cos (cos x) = cos − 2 cos−1 (cos x) = cos−1 (−1) The cosine function is negative in the 2nd and 3rd quadrants. 4π 2π and radians x= 3 3 Review Exercises 1. 129 x=π 1 − sin x = 1= √ √ 3 sin x → isolate sin x 3 sin x + sin x → simplify 1 = 2.73 sin x → solve 2.7321 sin x 1 = 2.7321 2.7321 0.3660 = sin x sin−1 (0.3660) = sin−1 sin x 0.3747radians = x Over the interval [0, π] the sine function is positive in the 2nd quadrant. x = π − .3747 x = 2.7669radians 2. 2 cos 3x − 1 = 0 2 cos 3x − 1 = 0 → isolate cos 3x 2 cos 3x 1 = 2 2 1 cos 3x = 2 ( ) 1 −1 −1 cos (cos 3x) = cos 2 1 cos 3x = 2 The interval [0, 2π] must be tripled since the equation has been solved for cos 3x, The interval is now [0, 6π]. To determine the values for x, each of these values must be divided by 3. π 5π 7π 11π 13π 17π , , , , , 3 3 3 3 3 3 π 5π 7π 11π 13π 17π x= , , , , , 9 9 9 9 9 9 3x = 3. 130 2 sec2 θ − tan4 θ = −1 2 sec2 θ − tan4 θ = −1 → sec2 θ = 1 + tan2 θ 2(1 + tan2 θ) − tan4 θ = −1 → expand 2 + 2 tan2 θ − tan4 θ = −1 → simplify 2 + 2 tan2 θ − tan4 θ + 1 = 0 → simplify − tan4 θ + 2 tan2 θ + 3 = 0 → ÷(−1) tan4 θ − 2 tan2 θ − 3 = 0 → factor (tan4 θ + 1)(tan2 θ − 3) = 0 → solve Then Or tan θ + 1 = 0 tan2 θ − 3 = 0 tan2 θ = −1 √ √ tan2 θ = −1 tan2 θ = 3 √ √ tan2 θ = 3 √ tan θ = ± 3 2 Does Not Exist √ tan−1 (tan θ) = tan−1 (± 3) For all real values of θ π π θ = + πk and θ = − + πk where k is any int eger 3 3 4. sin2 x − 2 = cos 2x sin2 x − 2 = cos 2x → cos 2x = 1 − 2 sin2 x sin2 x − 2 = 1 − 2 sin2 x → simplify sin2 x + 2 sin2 x = 1 + 2 → simplify 3 sin2 x = 3 → solve 3 sin2 x 3 = → solve 3 3 sin2 x = 1 √ √ sin2 x = ± 1 sin x = 1 sin x = −1 Over the interval 0◦ ≤ x < 360◦ 131 sin−1 (sin x) = sin−1 (1) x = 90◦ sin−1 (sin x) = sin−1 (1) x = 270◦ Solving Trigonometric Equations Using Inverse Notation Review Exercises 1. To solve y = π − arc sec 2x for x, the restricted range of arcsecant must be considered. y = π − arc sec 2x y = π − arc sec 2x → isolate arc sec 2x arc sec 2x = π − y 2x = sec(π − y) 1 x = − sec y 2 sec(π − y) = − sec y Since the values of arc sec 2x are restricted, so are the values of y. 2. To determine the value of sin(cot−1 (1)), the special triangles may be used or technology may be used. 1 (1) = 45◦ tan ( √ ) √ 1 2 2 ◦ = sin(45 ) = √ = √ 2 2 2 cot−1 (1) = Or sin 45◦ = 0.7071 → using techno log y 3. 5 cos x − 5 cos x − √ √ 2 = 3 cos x 2 = 3 cos x → isolare cos x √ 5 cos x − 3 cos x = 2 → simplify √ 2 cos x = 2 → simplify √ 2 cos x 2 = → simplify 2 2 132 cos x = √ 2 2 → The graph of the cosine function is one-to-one over the interval [0.π]. If the interval is restricted to [0.π], the arccosine of both sides of the equation would give an acceptable result. −1 cos −1 (cos x) = cos x= (√ ) 2 2 π which is within the restricted range of [0, π]. 4 However, this is the reference angle and the cosine function is also positive in the 4th quadrant. In this quadrant, the result would be x = 2π − π4 = 7π 4 which is within the interval [0, 2π]. To include all real solutions which would repeat every 2π units, the solutions for x could be expressed as x = π4 + 2πk where k is any int eger 4. sec θ − √ √ 2=0 2 = 0 → isolate sec θ √ 1 sec θ = 2 → sec θ = cos θ 1 cos θ = √ 2 sec θ − The graph of the cosine function is one-to-one over the interval [0, π]. If the interval is restricted to [0, π], the arccosine of both sides of the equation would give an acceptable result. −1 cos −1 (cos θ) = cos ( 1 √ 2 ) θ = 45◦ However, this is the reference angle and the cosine function is also positive in the 4th quadrant. In this quadrant, the result would be x = 360◦ − 45◦ = 315◦ which is within the interval 0◦ ≤ θ < 360◦ . To include all real solutions which would repeat every 360◦ , the solutions for x could be expressed as x = 45◦ + 360◦ k and where k is any integer and x = 315◦ + 360◦ k where k is any integer. Review Exercises 1. To solve i = Im [sin(wt + α) cos φ + cos(wt + α) sin φ] for t, the sum formula for sine must be applied. 133 i = Im [sin(wt + α) cos φ + cos(wt + α) sin φ] i = Im [sin(wt + α) cos φ + cos(wt + α) sin φ] → sin(a + b) = sin a cos b + cos a sin b → a = wt + α and b = φ i = Im [sin(wt + α) + (φ)] → simplify ( 1 w 1 w i Im ( 1 i w Im ( i 1 w Im ( i −1 sin Im ( i −1 sin Im i Im [sin(wt + α) + (φ)] = → simplify Im Im i I m [sin(wt + α) + (φ)] = → simplify Im I m i = sin(wt + α) + (φ) → simplify Im i − α − φ = sin wt + α − a + φ − φ → simplify Im ) − α − φ = sin wt → ÷(w) ) sin wt −α−φ = → simplify w ) − α − φ = sin t → solve ) − α − φ = sin−1 (sin t) ) −α−φ =t Review Exercises 1. Solving the following equation will not produce a numerical answer but it will result in an expression that is equal to theta. 134 I = I0 sin 2θ cos 2θ I = I0 sin 2θ cos 2θ → I0 I I0 sin 2θ cos 2θ = → simplify I0 I0 I = sin 2θ cos 2θ → ×(2) I0 ( ) I 2 = 2(sin 2θ cos 2θ) → simplify I0 2I = sin 4θ → solve I0 ( ) 2I sin−1 = sin−1 (sin 4θ) → simplify I0 ( ) 2I sin−1 = 4θ → ÷(4) I0 ( ) 2I sin−1 = 4θ → ÷(4) I0 ( ) ( ) 2I 4θ 1 sin−1 = 4 I0 4 ( ) ( ) 1 2I sin−1 =θ 4 I0 2. At ﬁrst glance, it seems that the diagram does not provide enough information. In order to obtain the answer, various values for theta will have to be substituted into the volume formula to determine when the maximum volume occurs. This question would be a great group activity. The volume of the trough is 10 times the area of the end of the trough. The end of the trough consists (of two identical right triangles. The ) area of each triangle is 21 (sin θ)(cos θ). The area of both triangles is 2 12 (sin θ)(cos θ) = (sin θ)(cos θ). The area of the rectangle is (1)(cos θ). The angles of the rectangle are 90◦ and the angles of the right triangles must be less than 90◦ . Therefore, the values of theta that must be considered are 0 ≤ θ ≤ π2 . The formula for the total volume of the trough is: V = 10(sin θ cos θ + cos θ) or V = 10(cos θ)(sin θ + 1) As values for theta are substituted into the formula, the calculated results must be recorded. The maximum volume is 13 ft3 and occurs when θ = π6 (30◦ ). 135 136 Chapter 5 TE Triangles and Vectors - Solution Key 5.1 Triangles and Vectors The Law of Cosines Review Exercises: 1. a) Using the two given sides and the included angle, the Law of Cosines must be used to calculate the length of side a. b) Using the lengths of the three given sides, the Law of Cosines must be used to calculate the measure of each of the three angles of △IRT . c) Using the two given sides and the included angle of △P LM the Law of Cosines must be used to calculate the length of side l(P M ). d) Using the lengths of the three given sides, the Law of Cosines must be used to determine the measure of the two remaining angles - ∠R and ∠D. e) Using the two given sides and the included angle, the Law of Cosines must be used to calculate the length of side b. f) Using the lengths of the three given sides, the Law of Cosines must be used to calculate the measure of each of the three angles of △CDM . 137 2. Given: ∠A = 50◦ , b = 8, c = 11 The length of side a can be determined by using the Law of Cosines. a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → ∠A = 50◦ , b = 8, c = 11 a2 = (8)2 + (11)2 − 2(8)(11) cos 50◦ → simplify This can be entered into the calculator, as shown, in one step. Press enter when complete. √ a2 = 71.8693807 → both sides √ √ √ a2 = 71.8693807 → both sides a ≈ 8.48 units b) Given: i = 11, r = 7, t = 6 138 The largest angle is across from the longest side. Therefore, determine the measure of ∠I using the Law of Cosines. r 2 + t2 − i2 2rt r 2 + t2 − i2 = → i = 11, r = 7, t = 6 2rt (7)2 + (6)2 + (11)2 → express answer as a fraction = 2(7)(6) −36 = → divide 84 = −0.4286 → A negative indicates that the angle is greater than 90◦ . cos ∠I = cos ∠I cos ∠I cos ∠I cos ∠I −1 cos (cos ∠I) = cos−1 (−0.4286) ∠I ≈ 115.4◦ c) Given: ∠L = 79.5◦ , m = 22.4, p = 13.7 l2 = m2 + p2 − 2mp cos L l2 = m2 + p2 − 2mp cos L → ∠L = 79.5◦ , m = 22.4, p = 13.7 l2 = (22.4)2 + (13.7)2 − 2(22.4)(13.7) cos(79.5◦ ) → simplify √ l2 = 577.6011 → both sides √ √ i2 = 577.6011 l ≈ 24.03 units 139 d) Given: d = 12.8, q = 17, r = 18.6, ∠Q = 62.4◦ The smallest angle is across from the shortest side. Therefore, determine the measure of ∠D using the Law of Cosines. q 2 + r2 − d2 2qr q 2 + r2 − d2 = → d = 12.8, q = 17, r = 18.6 2qr (17)2 + (18.6)2 − (12.8)2 = → simplify 2(17)(18.6) 471.12 = → divide 632.4 = 0.7450 cos ∠D = cos ∠D cos ∠D cos ∠D cos ∠D −1 cos (cos ∠D) = cos−1 (0.7450) ∠D ≈ 41.8◦ e) Given: d = 43, e = 39, ∠B = 67.2◦ 140 b2 = d2 + e2 − 2de cos B b2 = d2 + e2 − 2de cos B → d = 43, e = 39, ∠B = 67.2◦ b2 = (43)2 + (39)2 − 2(43)(39) cos(67.2◦ ) → simplify √ b2 = 2070.2727 → both sides b ≈ 45.5 units f) Given: c = 9, d = 11, m = 13 The second largest angle is across from the second longest side. Therefore, determine the measure of ∠D using the Law of Cosines. c2 + m2 − d2 2cm c2 + m2 − d2 = → c = 9, d = 11, m = 13 2cm (9)2 + (13)2 − (11)2 = → simplify 2(9)(13) 129 → divide = 234 = 0.5513 cos ∠D = cos ∠D cos ∠D cos ∠D cos ∠D −1 cos (cos ∠D) = cos−1 (0.5513) ∠D ≈ 56.5◦ 3. Given △CIR with c = 63, i = 52, r = 41.9. The Law of Cosines may be used to determine the measure of two of the angles and then the third can be determined by subtracting their sum from 180◦ . 141 i2 + r2 − c2 2ir i2 + r2 − c2 = → c = 63, i = 52, r = 41.9 2ir (52)2 + (41.9)2 − (63)2 = → simplify 2(52)(41.9) 490.61 = → divide 4357.6 = 0.1123 cos ∠C = cos ∠C cos ∠C cos ∠C cos ∠C −1 cos (cos ∠C) = cos−1 (0.1123) ∠C ≈ 83.5◦ c2 + r2 − i2 2cr c2 + r2 − i2 = → c = 63, i = 52, r = 41.9 2cr 2 (63) + (41.9)2 − (52)2 → simplify = 2(63)(41.9) 3020.61 = → divide 5279.4 = 0.5721 cos ∠I = cos ∠I cos ∠I cos ∠I cos ∠I −1 cos (cos ∠I) = cos−1 (0.5721) ∠I ≈ 55.1◦ ∠R ≈ 180◦ − (83.5◦ + 55.1◦ ) ∠R ≈ 41.4◦ 4. There are many ways to determine the length of AD. One way is to simply apply the trigonometric ratios. In △BCD: adj hyp x cos(37.4◦ ) = 14.2 x 0.7944 = 14.2 x (14.2)0.7944 = (14.2) 14.2 11.3 units ≈ x cos ∠C = 142 AD = AC − CD AD = 15 − 11.3 AD ≈ 3.7 units 5. In △HIK → HI = 6.7, IK = 5.2, ∠HIK = 96.3◦ . The Law of Cosines may be used to determine the length of HK. i2 = h2 + k 2 − 2hk cos I i2 = h2 + k 2 − 2hk cos I → HI(k) = 6.7, IK(h) = 5.2, ∠HIK(∠I) = 96.3◦ i2 = (5.2)2 + (6.7)2 − 2(5.2)(6.7) cos(96.3◦ ) → simplify √ i2 = 79.5763 → both sides √ √ i2 = 79.5763 i = 8.9 units 6. a) In △ABC → a = 20.9, b = 17.6, c = 15. The Law of Cosines may be used to conﬁrm the measure of ∠B. a2 + c2 − b2 2ac a2 + c2 − b2 = → a = 20.9, b = 17, c = 15 2ac (20.9)2 + (15)2 − (17.6)2 = → simplify 2(20.9)(15) 352.05 = → divide 627 = 0.5615 cos ∠B = cos ∠B cos ∠B cos ∠B cos ∠B −1 cos (cos ∠B) = cos−1 (0.5615) ∠B ≈ 55.8◦ △ABC is drawn accurately. b) In △DEF → d = 16.8, e = 24, f = 12. The Law of Cosines may be used to conﬁrm the measure of ∠D. 143 e2 + f 2 − d2 2ef 2 e + f 2 − d2 = → d = 16.8, e = 24, f = 12 2ef (24)2 + (12)2 − (16.8)2 → simplify = 2(4)(12) 437.76 = → divide 576 = 0.76 cos ∠D = cos ∠D cos ∠D cos ∠D cos ∠D −1 cos (cos ∠D) = cos−1 (0.76) ∠D ≈ 40.5◦ The Law of Cosines may now be applied to determine the correct length of side d. d2 = e2 + f 2 − 2ef cos D d2 = e2 + f 2 − 2ef cos D → ∠D = 30◦ , e = 24, f = 12 d2 = (24)2 + (12)2 − 2(24)(12) cos(30◦ ) → simplify √ d2 = 221.1694 → both sides √ √ d2 = 221.1694 d = 14.9 units △DEF is not accurately drawn. The length of side d is oﬀ by approximately 16.8 − 14.9 = 1.9 units. 7. To determine how long the cell phone service will last, the distance must be calculated and then this distance will have to be divided by the speed of the vehicle. The Law of Cosines may be used to calculate the distance. d2 = e2 + f 2 − 2ef cos D d2 = e2 + f 2 − 2ef cos D → e = 31 m, f = 26 m, ∠D = 47◦ d2 = (31)2 + (26)2 − 2(31)(26) cos(47◦ ) → simplify √ d2 = 537.6186 → both sides √ √ d2 = 537.6186 d = 23.2 m To determine the length of time that the cell phone service will last, divide this distance by the speed of 45 mph 23.2 m ≈ 0.52 hours ≈ 31.2 minutes 45 m/h If the answer for the distance is not rounded to 23.2 m as well as the answer for the number of hours, then the cell phone service will last approximately 30.9 minutes. 144 23.18660483 = 0.5152578851 ≈ (0.5152578851)(60) ≈ 30.9 minutes 45 b) 23.2 m 35 m/h ≈ 0.66 hours ≈ 39.6 minutes If the speed is reduced to 35 mph, the cell phone service will last for approximately 39.6 minutes which is 8.4 minutes longer. Or 23.18660483 = 0.6624744237 ≈ (0.6624744237)(60) ≈ 39.7 minutes 35 In this case, the cell phone service will last 8.8 minutes longer. 8. a) 145 a2 + c2 − b2 2ac a2 + c2 − b2 = → a = 306, b = 194.1, c = 183 2ac 2 (306) + (183)2 − (194.1)2 = → simplify 2(306)(183) 89450.19 = → divide 111996 = 0.7687 cos ∠B = cos ∠B cos ∠B cos ∠B cos ∠B −1 cos (cos ∠B) = cos−1 (07987) ∠B ≈ 37◦ The dock forms an angle of 37◦ with the two buoys. b) a2 + c2 − b2 2ac a2 + c2 − b2 = → a = 329, b = 207, c = 183 2ac (329)2 + (183)2 − (207)2 = → simplify 2(329)(183) 98881 = → divide 120414 = 0.8212 cos ∠B = cos ∠B cos ∠B cos ∠B cos ∠B −1 cos (cos ∠B) = cos−1 (0.8212) ∠B ≈ 34.8◦ If the distance from the second buoy to both the dock and the ﬁrst buoy is increased, the dock makes a smaller angle with the two buoys. The angle is 34.8◦ and is 2.2◦ smaller. 9. In △BCD, the Law of Cosines may be used to determine the length of DC (b) and then again to calculate the measure of ∠C. To determine the length of AB, the Law of Cosines can be used once again with △ABC. In △BCD: 146 b2 = c2 + d2 − 2cd cos B b2 = c2 + d2 − 2cd cos B → c = 32.6, d = 51.4, ∠B = 27◦ b2 = (32.6)2 + (51.4)2 − 2(32.6)(51.4) cos(77◦ ) → simplify √ b2 = 718.7077 → both sides √ √ b2 = 718.7077 b = 26.8 feet b2 + d2 − c2 2bd b2 + d2 − c2 = → b = 26.8, c = 32.6, d = 51.4 2bd (26.8)2 + (51.4)2 − (32.6)2 → simplify = 2(26.8)(51.4) 2297.44 = → divide 2755.04 = 0.8339 cos ∠C = cos ∠C cos ∠C cos ∠C cos ∠C −1 cos (cos ∠C) = cos−1 (0.8339) ∠C ≈ 33.5◦ In △ABC, a = 51.4, b = 37.3 + 26.8 = 64.1, ∠C = 33.5◦ . c2 = a2 + b2 − 2ab cos C c2 = a2 + b2 − 2ab cos C → a = 51.4, b = 64.1, ∠C = 33.5◦ c2 = (51.4)2 + (64.1)2 − 2(51.4)(64.1) cos(33.5◦ ) → simplify √ c2 = 1255.8961 → both sides √ √ c2 = 1255.8961 c ≈ 35.4 feet The length of AB is not 34.3 feet. It is 35.4 feet. 147 10. a) To determine the distance that the ball is from the hole, use the Law of Cosines to ﬁnd the length of side a. In △ABC → b = 329, c = 235, ∠A = 9◦ a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → b = 329, c = 235, ∠A = 9◦ a2 = (329)2 + (235)2 − 2(329)(235) cos(9◦ ) → simplify √ a2 = 10739.7519 → both sides √ √ a2 = 10739.7519 a ≈ 103.6 yds a ≈ 103.6 yards b) No solution. 11. There answers to this question are numerous. Below is one example of a possible solution. Three towns, A, B, and C respectively, are separated by distances that form a triangle. Town A is 127 miles from Town B and Town B is 210 miles from Town C. If the angle formed at Town B is 17◦ , calculate the number miles you would have to travel to complete a round trip that Begins at Town A. To answer this problem, the distance between Town A and Town C must be determined. Then the three distances must be added to determine the length of a round trip. 148 b2 = a2 + c2 − 2ac cos B b2 = a2 + c2 − 2ac cos B → a = 127, c = 210, ∠B = 17◦ b2 = (127)2 + (210)2 − 2(127)(210) cos(17◦ ) → simplify √ b2 = 9219.7043 → both sides √ √ b2 = 9219.7043 b ≈ 96.0 miles The distance you would travel to complete a round trip that begins in Town A is 127 + 210 + 96 = 433 miles. 12. There answers to this question are numerous. Below is one example of a possible solution. Given △ABC, calculate the area of the triangle to the nearest tenth. In △ABC, the Law of Cosines may be used to calculate the measure of ∠A or ∠C. b2 + c2 − a2 2bc b2 + c2 − a2 cos ∠A = → a = 27, b = 39, c = 15 2bc (39)2 + (15)2 − (27)2 cos ∠A = → simplify 2(39)(15) 1017 → divide cos ∠A = 1170 cos ∠A = 0.8692 cos ∠A = cos−1 (cos ∠A) = cos−1 (0.8692) ∠A ≈ 29.6◦ Use the sine ratio to calculate the height of the altitude. 149 opp hyp x ◦ sin 29.6 = 15 x 0.4939 = 15 ( sin ∠A = (15)(0.4939) = ( 15) 7.4 inches ≈ x The area of the triangle is 1 b·h 2 1 b · h → b = 39, h = 7.4 2 1 Area = (39) · (7.4) → simplify 2 Area = x) 1 5 Area = 144.3 in2 13. This question is similar to the one above. The additional step is to divide the area by 42000 ft2 to determine the number of acres of land. In △ABC, the Law of Cosines may be used to calculate the measure of ∠A or ∠C. b2 + c2 − a2 2bc b2 + c2 − a2 cos ∠A = → a = 600, b = 850, c = 300 2bc (850)2 + (300)2 − (600)2 cos ∠A = → simplify 2(850)(300) 452500 → divide cos ∠A = 510000 cos ∠A = 0.8873 cos ∠A = cos−1 (cos ∠A) = cos−1 (0.8873) ∠A ≈ 27.5◦ Use the sine ratio to calculate the height of the altitude. 150 opp hyp x sin 27.5◦ = 300 x 0.4617 = 300 ( x ) (300)(0.4617) = ( 300) 300 sin ∠A = The area of the triangle is 1 b·h 2 1 b · h → b = 850, h = 138.5 2 1 Area = (850) · (138.5) → simplify 2 Area = Area = 58862.5 ft2 138.5 feet ≈ x # of acres = 58862.5 ≈ 1.4 acres 42000 14. To determine the area of this quadrilateral, the area of triangles △ABC and △BCD, must be determined by using the Law of Cosines and the formula Area = 21 b · h. The area of each triangle must then be added to obtain the total area of the farm plot. In △ABC → a = 2200, b = 2400, c = 2100. The Law of cosines may be used to determine the measure of one of the angles of the triangle. a2 + c2 − b2 2ac a2 + c2 − b2 = → a = 2200, b = 2400, c = 2100 2ac (2200)2 + (2100)2 − (2400)2 = → simplify 2(2200)(2100) 3490000 = → divide 9240000 = 0.3777 cos ∠B = cos ∠B cos ∠B cos ∠B cos ∠B −1 cos (cos ∠B) = cos−1 (0.3777) ∠B ≈ 67.8◦ The length of the altitude drawn from A to BC can be calculated by using the sine ratio. 151 opp hyp x ◦ sin 67.8 = 2100 x 0.9259 = 2100 ( x ) (2100)(0.9259) = ( 2100) 2100 1944.4 feet ≈ x 1 Area = b · h → b = 2200, h = 1944.4 2 1 Area = (2200)(1944.4) → simplify 2 Area = 2, 138, 840 ft2 sin ∠B = In △BCD → b = 3000, c = 3000, d = 2200. The Law of cosines may be used to determine the measure of one of the angles of the triangle. b2 + d2 − c2 2bd b2 + d2 − c2 = → b = 3000, c = 3000, d = 2200 2bd (3000)2 + (2200)2 − (3000)2 = → simplify 2(3000)(2200) 4840000 = → divide 13200000 = 0.3667 cos ∠C = cos ∠C cos ∠C cos ∠C cos ∠C −1 cos (cos ∠C) = cos−1 (0.3667) ∠C ≈ 68.5◦ The length of the altitude drawn from D to BC can be calculated by using the sine ratio. opp hyp x ◦ sin 68.5 = 3000 x 0.9304 = 300 ( x ) (3000)(0.9304) = ( 3000) 3000 2791.3 feet ≈ x 1 b · h → b = 2200, h = 2791.3 2 1 Area = (2200) · (2791.3) 2 sin ∠C = Area = Area ≈ 3, 070, 430 ft2 The total area of the quadrilateral farm plot is approximately: 2, 138, 840 ft2 + 3, 070, 430 ft2 = 5, 209, 270 ft2 15. To determine the length of the cable at each of the reaches, the Law of Cosines can be used. 152 a) b2 = a2 + c2 − 2ac cos B b2 = a2 + c2 − 2ac cos B → a = 20, c = 4, ∠B = 17◦ b2 = (20)2 + (4)2 − 2(20)(4) cos(17◦ ) → simplify √ b2 = 262.9912 → both sides b ≈ 16.2 m The cable is approximately 16.2 m long at the crane’s lowest reach. b) b2 = a2 + c2 − 2ac cos B b2 = a2 + c2 − 2ac cos B → a = 20, c = 4, ∠B = 82◦ b2 = (20)2 + (4)2 − 2(20)(4) cos(82◦ ) → simplify √ b2 = 393.7323 → both sides b ≈ 19.8 m The cable is approximately 19.8 m long at the crane’s highest reach. 16. To solve this problem the Law of Cosines will have to used to determine the length of AB and then used gain to calculate the measure of ∠AEH. e2 = a2 + b2 − 2ab cos E e2 = a2 + b2 − 2ab cos E → a = 4, b = 21, ∠E = 120◦ e2 = (4)2 + (21)2 − 2(4)(21) cos(120◦ ) → simplify √ e2 = 541 → both sides e ≈ 23.3 cm The length of AB is reduced by 5 cm. when the ﬂuid is pumped out of the cylinder. As a result, The length of 23.3 − 5.0 = 18.3 cm must be used to calculate the measure of ∠AEH. a2 + b2 − e2 2ab a2 + b2 − e2 = → a = 4, b = 21, e = 18.3 2ab 2 2 (4) + (21) − (18.3)2 → simplify = 2(4)(21) 122.11 = → divide 168 = 0.7268 cos ∠E = cos ∠E cos ∠E cos ∠E cos ∠E −1 cos (cos ∠E) = cos−1 (0.7268) ∠E ≈ 43.4◦ 153 Area of a Triangle Review Exercises: 1. a) In △COM , Pythagorean Theorem can be used to determine the height of the altitude OF . Then, the length of the base can be calculated by adding the given lengths of CF and F M . With these two measurements, the formula A = 21 b · h may be used to obtain the area of the triangle. b) The area of △CEH can be calculated by applying Heron’s Formula since the length of the each side of the triangle is given. c) In △AP H, the length of two sides is given as well as the measure of the included angle. The K = 21 bc sin A formula may be used to determine the area of the triangle. d) In △XLR, the tangent ratio can be used to determine the height of the altitude LX. Then, the length of the base can be calculated by adding the given lengths of RX and XE. With these two measurements, the formula A = 21 b · h may be used to obtain the area of the triangle. 2. a) In △COF (h)2 = (s1 )2 + (s2 )2 2 2 Base(CM ) = CF + F M ) 2 (5) = (3) + (s2 ) (CM ) = 3 + 8 = 11 25 − 9 = s √ √ 16 = s2 4 = s (OF) 2 1 b·h 2 1 A = (11) · (4) 2 A = 22 units2 A= b) √ s(s − a)(s − b)(s − c) √ 1 K = s(s − a)(s − b)(s − c) → s = (4.1 + 7.4 + 9.6) = 10.55 2 → c(a) = 9.6, e(b) = 4.1, h(c) = 7.4 √ K = 10.55(10.55 − 9.6)(10.55 − 4.1)(10.55 − 7.4) → simplify √ K = 203.6321438 K= K = 14.27 units2 154 c) 1 bc sin A 2 1 K = bc sin A → b(a) = (86.3), c(h) = 59.8, ∠P (A) = 103◦ 2 1 K = (86.3)(59.8) sin(103◦ ) → simplify 2 K = 2514.24 units2 K= d) opp adj x ◦ tan 41 = 11.1 x 0.8693 = 11.1 ( x ) (11.1)(0.8693) = ( 11.1) 11.1 9.6 ≈ x tan θ = Base(ER) = RX + XE (ER) = 11.1 + 18.9 = 30 1 b·h 2 1 A = (30) · (9.6) 2 A = 144 units2 A= 3. a) In △ABC, the area and the length of the base are given. To determine the length of the A = 12 b · h altitude, the formula must be used. b) In △ABC. The area and the lengths of two sides of the triangle are given. To determine the measure of the included angle, the formula K = 21 b · c sin A must be used. c) In △ABD, the formula A = 12 b · h, can be used to determine the length of the altitude. With this measurement calculated, the tangent function can be used to determine the length of CD. The formula A = 21 b · h can now be used to calculate the area of △ABC. 4. a) 1 b·h 2 1 A = b · h → A = 1618.98, b = 36.3 2 1 1618.98 = (36.3) · h → simplify 2 1618.98 = 18.15h → solve 18.15 1618.98 = h → solve 18.15 18.15 A= 89.2 units = h 155 b) 1 bc sin A 2 1 K = bc sin A → A = 387.6, b = 25.6, c = 32.9 2 1 387.6 = (25.6)(32.9) sin A → simplify 2 387.6 = 421.12 sin A → solve 387.6 421.12 = sin A → solve 421.12 421.12 0.9204 = sin A K= sin−1 (0.9204) = sin−1 (sin A) 67◦ ≈ ∠A c) 1 b·h 2 1 A = b · h → A = 16.96, b = 3.2 2 1 16.96 = (3.2) · h → simplify 2 16.96 = 1.6h → solve 1.6 16.96 = h 1.6 1.6 10.6 units = h A= In △BCD: opp adj x tan 49.6◦ = 10.6 tan ∠B = 1.1750 = x 10.6 (10.6)(1.1750) = ( 10.6) ( x ) → 12.5 units ≈ x 10.6 The total length of the base (AC) is 3.2 + 12.5 = 15.7 units. The area of △ABC is 1 b·h 2 1 A = b · h → b = 15.7, h = 10.6 2 1 A = (15.7) · (10.6) → simplify 2 A = 83.21 units2 A= 5. a) To determine the total area of the exterior of the Pyramid Hotel, Heron’s Formula should be used. The four sides are isosceles triangles so the area of one side can be multiplied by 4 to obtain the total area. 156 1 (a + b + c) 2 1 s = (a + b + c) → a = 375, b = 375, c = 590 2 1 s = (375 + 375 + 590) → simplify 2 s = 670 feet → one side s= Area of one side: √ s(s − a)(s − b)(s − c) √ K = s(s − a)(s − b)(s − c) → s = 670, a = 375, b = 375, c = 590 √ K = 670(670 − 375)(670 − 375)(670 − 590) → simplify √ √ K = 4664540000 → K= K = 68, 297.4 ft2 → one side Total Area: (4)68, 297.4 ft2 = 273, 189.6 ft2 b) The number of gallons of paint that are needed to paint the hotel is: 273, 189.6 ft2 = 10927.584 ≈ 109, 28 gallons 25 ft2 5.a) The three sides of the triangular section a have been given in the problem. Therefore, Heron’s Formula may be used to calculate the area of the section. 1 (a + b + c) 2 1 s = (a + b + c) → a = 8.2, b = 14.6, c = 16.3 2 1 s = (8.2 + 14.6 + 16.3) → simplify 2 s = 19.55 feet s= √ s(s − a)(s − b)(s − c) √ K = s(s − a)(s − b)(s − c) → s = 19.55, a = 8.2, b = 14.6, c = 16.3 √ K = 19.55(19.55 − 8.2)(19.55 − 14.6)(19.55 − 16.3) → simplify √ √ K = 3569.695594 → K= K = 59.7 ft2 The number of bundles of shingles that must be purchased is: 157 59.7 ÷ 33 1 = 1.791 ≈ 2 3 b) The shingles will cost (2)($15.45) = $30.90 c) The shingles that will go to waste are: 2 − 1.791 = .209 ( ) 1 (.209) 33 ≈ 6.97 ft2 3 7. a) To determine the area of the section of crops that need to be replanted, the formula K = may be used because the lengths of two sides and the included angle are known.. 1 2 bc sin A 1 bc sin A 2 1 K = bc sin A → b = 186, c = 205, ∠A = 148◦ 2 1 K = (186)(205) sin 148◦ → simplify 2 K ≈ 10, 102.9 yd2 K= b) 1 bc sin A 2 1 K = bc sin A → b = 186, c = 288, ∠A = 148◦ 2 1 K = (186)(288) sin 148◦ → simplify 2 K ≈ 14, 193.4 yd2 K= The increase in the area that must be replanted is: 10, 102.9 yd2 14, 193.4 yd2 − 10, 102.9 yd2 = 4090.5 yd2 8. The length of one side of each triangle can be determined by using the formula K = 12 bc sin A. The third side of each triangle can be found by using the Law of Cosines. The perimeter of the quadrilateral can then be determined by adding the lengths of the sides. △DEG: 158 1 dg sin E 2 1 K = dg sin E → K = 56.5, d = 13.6, ∠E = 39◦ 2 1 56.5 = (13.6)g sin(39◦ ) → simplify 2 1 56.5 = (13.6)g(0.6293) → simplify 2 56.5 = 4.27924g → solve ( (( ( 56.5 4.27924 = (( (g 4.27924 ( 4.27924 13.2 units ≈ g K= e2 = d2 + g 2 − 2dg cos E e2 = d2 + g 2 − 2dg cos E → d = 13.6, g = 13.2, ∠E = 39◦ e2 = (13.6)2 + (13.2)2 − 2(13.6)(13.2) cos(39◦ ) → simplify √ e2 = 80.1735 → both sides √ √ e2 = 80.1735 e ≈ 9.0 units △EF T 1 ef sin G 2 1 K = ef sin G → K = 84.7, f = 13.6, ∠G = 60◦ 2 1 84.7 = e(13.6) sin(60◦ ) → simplify 2 1 84.7 = e(13.6)(0.8660) → simplify 2 84.7 = 5.8888e → solve 84.7 5.8888 = e 5.8888 5.8888 14.4 units ≈ e K= g 2 = e2 + f 2 − 2ef cos G g 2 = e2 + f 2 − 2ef cos G → e = 14.4, f = 13.6, ∠G = 60◦ g 2 = (14.4)2 + (13.6)2 − 2(14.4)(13.6) cos(60◦ ) → simplify √ g 2 = 196.48 → both sides √ √ g 2 = 196.48 g ≈ 14.0 units 159 The perimeter of the quadrilateral is approximately 13.2 + 9.0 + 14.4 + 14.0 = 50.6 units. 9. In the following triangle, Pythagorean Theorem can be used to determine the length of the altitude BD. Then the formula A = 21 b · h can be used to determine the length of the base AC. The diﬀerence between the base and AD is the length of DC. (h)2 = (s1 )2 + (s2 )2 (h)2 = (s1 )2 + (s2 )2 → h = 16.2, s1 = 14.4 (16.2)2 = (14.4)2 + (s2 )2 → simplify (16.2)2 − (14.4)2 = (s2 )2 → simplify √ 55.08 = (s2 )2 → both sides √ √ 55.08 = (s2 )2 7.4 ≈ s The altitude of the triangle is approximately 7.4 units. 1 b·h 2 1 A = b · h → A = 232.96, h = 7.4 2 1 232.96 = b · (7.4) → simplify 2 232.96 = 3.7b → solve 232.96 3.7 = b → solve 3.7 3.7 63.0 units ≈ b A= DC = AC − AD DC = 63.0 − 14.4 = 48.6 units 10. To show that in any triangle DEF, d2 + e2 + f 2 = 2(ef cos D + df cos E + de cos F ) the Law of Cosines for ﬁnding the length of each side, d, e and f will have to be used and the sum of these will have to be simpliﬁed. 160 d2 = e2 + f 2 − 2ef cos D e2 = d2 + f 2 − 2df cos E f 2 = d2 + e2 − 2de cos F d2 + e2 + f 2 = (e2 + f 2 − 2ef cos D) + (d2 + f 2 − 2df cos E) + (d2 + e2 − 2de cos F ) → simplify d2 + e2 + f 2 = 2d2 + 2e2 + 2f 2 − 2ef cos D − 2df cos E − 2de cos F → simplify d2 + e2 + f 2 = 2d2 + 2e2 + 2f 2 − 2ef cos D − 2df cos E − 2de cos F → common factor d2 + e2 + f 2 = 2(d2 + e2 + f 2 ) − 2(ef cos D + df cos E + de cos F ) → simplify d2 + e2 + f 2 − 2(d2 + e2 + f 2 ) = −2(ef cos D + df cos E + de cos F ) → simplify −(d2 + e2 + f 2 ) = −2(ef cos D + df cos E + de cos F ) → ÷(−1) d2 + e2 + f 2 = 2(ef cos D + df cos E + de cos F ) The Law of Sines Review Exercises: 1. a) This situation represents ASA. b) This situation represents AAS. c) This situation represents neither ASA nor AAS. The measure of the 3 angles is given. d) This situation represents ASA. e) This situation represents AAS. f) This situation represents AAS. 2. In all of the above cases, the length of the side across from an angle can be determined. As well, the measure of an angle can be determined. 3. a) To determine the length of side a, the measure of ∠B must be calculated ﬁrst. This can be done by adding the two given angles and subtracting their sum from 180◦ . Then the Law of Sines can be used to determine the length of side a. ∠B = 180◦ − (11.7◦ + 23.8◦ ) ∠B = 180◦ − (35.5◦ ) ∠B = 144.5◦ Law of Sines: 161 a b = sin A sin B a b = → b = 16, ∠A = 11.7◦ , ∠B = 144.5◦ sin A sin B a 16 = → simplify sin(11.7◦ ) sin(144.5◦ ) a(sin(144.5◦ )) = 16(sin(11.7◦ )) → simplify a(0.5807) = 16(0.2028) → solve 0.5807a = 3.2448 → solve 0.5807a 3.2448 = 0.5807 0.5807 a ≈ 5.6 units b) To determine the length of side d, the Law of Sines must be applied. e d = sin E sin D e d = → e = 214.9, ∠D = 39.7◦ , ∠E = 41.3◦ sin E sin D 214.9 d = → simplify ◦ sin(41.3 ) sin(39.7◦ ) 214.9(sin(39.7◦ )) = d(sin(41.3◦ )) → simplify 214.9(0.6388) = d(0.6600) → simplify 137.2781 = (0.6600)d → solve 137.2781 (0.6600) = d → solve 0.6600 0.6600 208.0 units ≈ d c) Cannot determine the length of side i. There is not enough information provided. d) To determine the length of side l, the measure of ∠K must be calculated ﬁrst. This can be done by adding the two given angles and subtracting their sum from 180◦ . Then the Law of Sines can be used to determine the length of side l. ∠K = 180◦ − (16.2◦ + 40.3◦ ) ∠K = 180◦ − (56.5◦ ) ∠K = 123.5◦ 162 k l = sin K sin L k l = → k = 6.3, ∠K = 123.5◦ , ∠L = 40.3◦ sin K sin L 6.3 l = → simplify sin(123.5◦ ) sin(40.3◦ ) 6.3(sin(40.3◦ )) = l(sin(123.5◦ )) → simplify 6.3(0.6468) = (0.8339)l → simplify 4.0748 = (0.8339)l → solve (0.8339) 4.0748 = l → solve 0.8339 0.8339 4.9 units ≈ l e) To determine the length of side o, the Law of Sines must be applied. o m = sin O sin M o m = → m = 15, ∠O = 9◦ , ∠M = 31◦ sin O sin M o 15 = → simplify sin(9◦ ) sin(31◦ ) o(sin(31◦ )) = 15(sin(9◦ )) → simplify o(0.5150) = 15(0.1564) → simplify (0.5150)o = 2.346 → solve (0.5150) 2.346 o = 0.5150 → solve 0.5150 o ≈ 4.6 units f) To determine the length of side q, the Law of Sines must be applied. q r = sin Q sin R q r = → r = 3.62, ∠Q = 127◦ , ∠R = 21.8◦ sin Q sin R q 3.62 = → simplify ◦ sin(127 ) sin(21.8◦ ) q(sin(21.8◦ )) = 3.62(sin(127◦ )) → simplify q(0.3714) = 3.62(0.7986) → simplify (0.3714)q = 2.8909 → solve 2.8909 (0.3714) q = 0.3714 → solve 0.3714 q ≈ 7.8 units 4. To determine the length of side h, the Law of Sines may be used. The Law of Sines may also be used to determine the length of side g after the measure of ∠G is calculated. This can be done by adding the two given angles and subtracting their sum from 180◦ . 163 ∠G = 180◦ − (62.1◦ + 21.3◦ ) ∠G = 180◦ − (83.4◦ ) ∠G = 96.6◦ Side h i h = sin H sin I h i = → i = 108, ∠H = 62.1◦ , ∠I = 21.3◦ sin H sin I 108 h = → simplify ◦ sin(62.1 ) sin(21.3◦ ) h(sin(21.3◦ )) = 108(sin(62.1◦ )) → simplify h(0.3633) = 108(0.8838) → simplify (0.3633)h = 95.450 → solve (0.3633) 95.450 h = 0.3633 → solve 0.3633 h ≈ 262.7 units Side g g i = sin G sin I g i = → i = 108, ∠G = 96.6◦ , ∠I = 21.3◦ sin G sin I 108 g = → simplify sin(96.6◦ ) sin(21.3◦ ) g(sin(21.3◦ )) = 108(sin(96.6◦ )) → simplify g(0.3633) = 108(0.9934) → simplify (0.3633)g = 107.287 → solve 107.287 (0.3633) g = 0.3633 → solve 0.3633 g ≈ 295.3 units 164 5. sin A sin B = a b b(sin A) = a(sin B) b(sin A) a(sin B) = b b b(sin A) a(sin B) = b b a(sin B) (sin A) = b B) a (sin (sin A) = B) (sin B) b (sin (sin A) a = (sin B) b 6. a) The Law of Cosines because the lengths of two sides and the included angle are given. b) The triangle is a right triangle so the assumption that would be made is that one of the Trigonometric ratios would be used to determine the length of side x. However, there is not enough information given to conclude which ratio to apply. c) Either the Law of Cosines or the Law of Sines could be used to calculate the measure of the angle. d) The Law of Sines would be used to determine the length of side x. 7. a) opp adj x ◦ tan(54 ) = 7.15 x 1.3764 = 7.15 ( x ) 1.3764(7.15) = ( 7.15) 7.15 9.84 units ≈ x opp adj x ◦ tan(67 ) = 9.84 x 2.3559 = 9.84 ( x ) 2.3559(9.84) = ( 9.84) 9.84 23.2 units ≈ x tan(54◦ ) = tan(67◦ ) = b) To determine the length of side x, the Law of Cosines can be used to determine the measure of the supplementary angle. This measurement can then be subtracted from 180◦ to calculate the measure of the corresponding angle and the Law of Sines can then be applied. 165 b2 + c2 − a2 2bc b2 + c2 − a2 cos ∠A = → a = 11.2, b = 12.6, c = 8.9 2bc (12.6)2 + (8.9)2 − (11.2)2 cos ∠A = → simplify 2(12.6)(8.9) 112.53 cos ∠A = → divide 224.28 cos ∠A = 0.5017 cos ∠A = cos−1 (cos ∠A) = cos−1 (0.5017) ∠A ≈ 59.9◦ Supplementary Angle: 180◦ − 59.9◦ = 120.1◦ . This is also the corresponding angle in the other triangle. c a = sin A sin C a c = → c = 8.9, ∠A = 120.1◦ , ∠C = 31◦ sin A sin C a 8.9 = → simplify sin(120.1◦ ) sin(31◦ ) a(sin(31◦ )) = 8.9(sin(120.1◦ )) → simplify a(0.5150) = 8.9(0.8652) → simplify (0.5150)a = 7.7 → solve (0.5150) 7.7 a = 0.5150 → solve 0.5150 a ≈ 15.0 units 8. There is not enough information given to complete this problem. 9. To determine the time that the driver must leave the warehouse, the total distance she travels and the length of time to travel the distance must be calculated. The distance between Stop B and Stop C can be determined by using the Law of Sines. The distance between Stop A and Stop C can also be determined by using the Law of Sines. The angle formed by the intersection of Stop C and Route 52 can be calculated by subtracting the sum of the other 2 angles from 180◦ . ∠C = 180◦ − (41◦ + 103◦ ) ∠C = 180◦ − (144◦ ) ∠C = 36◦ Distance between Stop B and Stop C (a) 166 a c = sin A sin C a c = → c = 12.3, ∠A = 41◦ , ∠C = 36◦ sin A sin C a 12.3 = → simplify ◦ sin(41 ) sin(36◦ ) a(sin(36◦ )) = 12.3(sin(41◦ )) → simplify a(0.5878) = 12.3(0.6561) → simplify (0.5878) 8.070 a= → solve 0.5878 0.5878 a ≈ 13.8 units Distance between Stop A and Stop C (b) b c = sin B sin C b c = → c = 12.3, ∠B = 103◦ , ∠C = 36◦ sin B sin C b 12.3 = → simplify sin(103◦ ) sin(36◦ ) b(sin(36◦ )) = 12.3(sin(103◦ )) → simplify b(0.5878) = 12.3(0.9744) → simplify b(0.5878) = 11.985 → solve 11.985 (0.5878) b = 0.5878 → solve 0.5878 b ≈ 20.4 miles The total distance the driver must travel is 1.1 + 12.3 + 20.4 + 13.8 + 1.1 = 48.7 miles miles ≈ 1.1 hours or 1 hour and To travel this distance at a speed of 45 mph will take the driver 48.7 45 mph 6 minutes. The driver must add to this time, the time needed to deliver each package. Now the total time is 1 hour 12 minutes. In order to return to the warehouse by 10:00 a.m., she must leave the warehouse at 8:48 a.m. 10. The information given in his problem is not suﬀicient to obtain an answer. If an angle of elevation increases, then the observer must be closer to the object. If this is the case, then the problem does not work. The Ambiguous Case Review Exercises: 1. a) 167 b sin A b sin A → b = 37, ∠A = 32.5◦ (37) sin(32.5◦ ) → simplify (37)(0.5373) → simplify (37)(0.5373) ≈ 19.9 Therefore a > b sin A and there will be two solutions. b) b sin A b sin A → b = 26, ∠A = 42.3◦ (26) sin(42.3◦ ) → simplify (26)(0.6730) → simplify (26)(0.6730) ≈ 17.5 Therefore a < b sin A and there are no solutions. c) b sin A b sin A → b = 18.2, ∠A = 47.8◦ (18.2) sin(47.8◦ ) → simplify (18.2)(0.7408) → simplify (18.2)(0.7408) ≈ 13.5 168 Therefore a = b sin A and there is one solution. d) b sin A b sin A → b = 4.2, ∠A = 51.5◦ (4.2) sin(51.5◦ ) → simplify (4.2)(0.7826) → simplify (4.2)(0.7826) ≈ 3.3 Therefore a > b sin A and there will be two solutions. 2. a) sin A sin B = a b sin B sin A = → ∠A = 32.5◦ , a = 26, b = 37 a b sin(32.5◦ ) sin B = → simplify 26 37 sin(32.5◦ )(37) = (26) sin B → simplify (0.5373)(37) = (26) sin B → simplify 19.8801 = (26) sin B → solve sin B 19.8801 (26) = → solve 26 2 6 0.7646 = sin B → solve sin−1 (0.7646) = sin−1 (sin B) 49.9◦ ≈ ∠B OR ∠B = 180◦ − 49.9◦ = 130.1◦ b) There is no solution as proven above in question 1. 169 c) sin A sin B = a b sin A sin B = → ∠A = 47.8◦ , a = 13.5, b = 18.2 a b sin B sin(47.8◦ ) = → simplify 13.5 18.2 sin(47.8◦ )(18.2) = (13.5) sin B → simplify (0.7408)(18.2) = (13.5) sin B → simplify 13.4826 = (13.5) sin B → solve sin B 13.4826 (13.5) = → solve 13.5 13.5 0.9987 = sin B → solve sin−1 (0.9987) = sin−1 (sin B) 87.1◦ ≈ ∠B d) sin A sin B = a b sin A sin B = → ∠A = 51.5◦ , a = 3.4, b = 4.2 a b sin(51.5◦ ) sin B = → simplify 3.4 4.2 (4.2) sin(51.5◦ ) = (3.4) sin B → simplify (4.2)(0.7826) = (3.4) sin B → simplify 3.2869 = (3.4) sin B → solve sin B 3.2869 (3.4) → solve = 3.4 3.4 0.9667 = sin B sin−1 (0.9667) = sin−1 (sin B) 75.2◦ ≈ ∠B OR ∠B = 180◦ − 75.2◦ = 104.8◦ 170 3. sin A sin C = a) c ( ( ) sin A sin C (ac) = (ac) → simplify a c ( ) ) ( sin A sin C (ac) = (a c) → simplify a c (c)(sin A) = (a)(sin C) → simplify cSinA − cSinC = aSinC − cSinC → common factor (c)(SinA − SinC) = (SinC)(a − c) → divide(cSinC) (Sin C)(a − c) (c)(SinA − SinC) = → simplify cSinC cSinC ( SinC)(a − c) (c)(SinA − SinC) = → simplify c SinC cSinC SinA − SinC (a − c) = SinC c 4. Given △ABC → a = 30 cm, c = 42 cm, ∠A = 38◦ . To calculate the measure of ∠C, the Law of Sines must be used. sin A sin C = a c sin A sin C = → a = 30 cm, c = 42 cm, ∠A = 38◦ a c sin(38◦ ) sin C = → simplify 30 42 sin C 0.6157 = → simplify 30 42 (0.6157)(42) = (30)(sin C) → simplify 25.8594 = 30 sin C → simplify sin C 25.8594 30 → simplify = 30 30 0.8620 = sin C sin−1 (0.8620) = sin−1 (sin C) 59.5◦ ≈ ∠C OR ∠C = 180◦ − 59.5◦ = 120.5◦ There are two solutions which results in two triangles. The sine function is positive in both the 1st and 2nd quadrant. The two possibilities are given above and both will satisfy the measure of the angle. Therefore, the length of side ‘b’ will depend upon its corresponding angle. ∠B = 180◦ − (38◦ + 59.5◦ ) ∠B = 180◦ − (97.5◦ ) ∠B = 82.5◦ OR 171 ∠B = 180◦ − (38◦ + 120.5◦ ) ∠B = 180◦ − (158.5◦ ) ∠B = 21.5◦ sin A sin B = a b sin A sin B = → a = 30 cm, ∠B = 82.5◦ , ∠A = 38◦ a b sin(82.5◦ ) sin(38◦ ) = → simplify 30 b 0.6157 0.9914 = → simplify 30 b (0.6157)(b) = (30)(0.9914) → simplify 0.6157b = 29.742 → solve 0.6157b 29.742 = → solve 0.6157 0.6157 b ≈ 48.3 cm sin A sin B = a b sin A sin B = → a = 30 cm, ∠B = 21.5◦ , ∠A = 38◦ a b sin(21.5◦ ) sin(38◦ ) = → simplify 30 b 0.6157 0.3665 = → simplify 30 b (0.6157)(b) = (30)(0.3665) → simplify 0.6157b = 10.995 → solve 0.6157b 10.995 = → solve 0.6157 0.6157 b ≈ 17.9 cm Two triangles exist: 5. If there is one solution, a = b sin A. In order for this to be true, the measure of ∠A must be calculated. 172 a = b sin A a = b sin A → a = 22, b = 31 22 = 31 sin A → solve 22 3 1 sin A = → solve 31 31 0.7097 = sin A sin−1 (0.7097) = sin−1 (sin A) 45.2◦ ≈ ∠A a) No solution means that a = b sin A. This will occur when ∠A is greater than 45.2◦ . b) One solution means that a = b sin A. This will occur when ∠A equals 45.2◦ . c) Two solutions mean that a = b sin A. This will occur when ∠A is less than 45.2◦ . 6. In the following triangle, the trigonometric ratios may be used to determine the measure of the required angles and sides or these may be used in conjunction with the Law of Cosines or the Law of Sines. △ACD opp sin ∠C = hyp 9.8 ◦ sin(42.6 ) = x 9.8 0.6769 = x ( 0.6769(x) = (x) △ABD opp hyp 9.8 sin ∠B = 13.7 sin ∠B = 9.8 x sin ∠B = 0.7153 ) sin−1 (sin ∠B) = sin−1 (0.7153) ∠B ≈ 45.7◦ 0.6769x = 9.8 9.8 0.6769 x= 0.6769 0.6769 9.8 0.6769 x = 0.6769 0.6769 x ≈ 14.5 units 173 ∠A = 180◦ − (42.6◦ + 45.7◦ ) ∠A = 180◦ − (88.3◦ ) ∠A = 91.7◦ △ABC a = b2 + c2 − 2bc cos A 2 a2 = b2 + c2 − 2bc cos A → b = 14.5, c = 13.7, ∠A = 91.7◦ a2 = (14.5)2 + (13.7)2 − 2(14.5)(13.7) cos(91.7◦ ) → simplify √ a2 = 409.7264 → both sides √ √ a2 = 409.7264 → simplify a ≈ 20.2 units The required measurements are: ∠A = 91.7◦ , ∠B = 45.7◦ , AC = 14.5 units, Bc = 20.2 units 7. To determine the measurements of the required sides and angles, the Law of Cosines, the Law of Sines, supplementary angles and the sum of the angles of a triangle must be used. Begin with △BED since the length of each side is given. Use the Law of Cosines to determine the measure of the angles and then apply the sum of the angles in a triangle to determine the third angle. Then continue until the measure of each angle has been calculated. △BED d2 + e2 − b2 2de d2 + e2 − b2 = → b = 7.6, d = 9.9, e = 10.2 2de (9.9)2 + (10.2)2 − (7.2)2 = → simplify 2(9.9)(10.2) 144.29 → divide = 201.96 = 0.7144 cos B = cos B cos B cos B cos B −1 cos −1 ∠D and∠BDC are supplementary angles ∴ ∠BDC = 180◦ − 65.7◦ = 114.3◦ (cos B) = cos (0.7144) ∠B ≈ 44.4◦ 174 b2 + d2 − e2 2bd b2 + d2 − e2 = → b = 7.6, d = 9.9, e = 10.2 2bd (7.6)2 + (9.9)2 − (10.2)2 = → simplify 2(7.6)(9.9) 51.73 = → divide 150.48 = 0.3438 cos E = cos E cos E cos E cos E −1 cos (cos E) = cos−1 (0.3438) ∠E ≈ 69.9◦ ∠E and∠BEA are supplementary angles ∴ ∠BEA = 180◦ − 69.9◦ = 110.1◦ ∠D = 180◦ − (44.4◦ + 69.9◦ ) ∠D = 180◦ − (114.3◦ ) ∠D = 65.7◦ In △CBD ∠B = 180◦ − (114.3◦ + 21.8◦ ) ∠B = 180◦ − (136.1◦ ) In △ABC ∠A = 180◦ − (109.6◦ + 21.8◦ ) ∠A = 180◦ − (131.4◦ ) In △ABE ∠B = 180◦ − (110.1◦ + 48.6◦ ) ∠B = 180◦ − (158.7◦ ) ∠B = 43.9◦ ∠A = 48.6◦ ∠B = 21.3◦ In △ABE, the length of AB is determined by using the Law of Sines. a e = sin A sin E a e = → a = 9.9, ∠A = 48.6◦ , ∠E = 110.1◦ sin A sin E 9.9 e = → simplify ◦ sin(48.6 ) sin(110.1◦ ) (9.9)(sin(110.1◦ )) = (sin(48.6◦ ))e → simplify 9.9(0.9391) = 0.7501e → simplify 9.2971 = 0.7501e → solve 0.7501e 9.2971 = → solve 0.7501 0.7501 12.4 units ≈ e(AB) In △BCD, the length of BC is determined by using the Law of Sines. 175 c d = sin C sin D c d = → c = 10.2, ∠C = 21.8◦ , ∠D = 114.3◦ sin C sin D 10.2 d = → simplify sin(21.8◦ ) sin(114.3◦ ) (sin(114.3◦ ))10.2 = (sin(21.8◦ ))d → simplify (0.9114)10.2 = (0.3714)d → simplify 9.2963 = (0.3714)d → solve (0.3714)d 9.2963 = → solve 0.3714 0.3714 25.0 units ≈ d(BC) In △BCD, the length of DC is determined by using the Law of Sines b c = sin C sin B c b = → c = 10.2, ∠C = 21.8◦ , ∠B = 43.9◦ sin C sin B 10.2 b = → simplify sin(21.8◦ ) sin(43.9◦ ) (sin(43.9◦ ))10.2 = (sin(21.8◦ ))b → simplify (0.6934)10.2 = (0.3714)b → simplify 7.0727 = (0.3714)b → solve (0.3714)b 7.0727 = → solve 0.3714 0.3714 19.0 units ≈ b(CD) In △ABC, the Law of Cosines may be used to calculate the length of side b (AC) b2 = a2 + c2 − 2ac cos B b2 = a2 + c2 − 2ac cos B → a = 25, c = 12.4, ∠B = 109.6◦ b2 = (25)2 + (12.4)2 − 2(25)(12.4) cos(109.6◦ ) → simplify √ b2 = 986.7400 → both sides √ √ b2 = 986.7400 → simplify b(AC) ≈ 31.4 units In △ABE, the length of AE is the diﬀerence between the length of AC and the sum of ED and CD. AE = AC − (ED + CD) AE = 31.4 − (7.6 + 19.0) AE = 4.8 units 176 The solutions are: a) BC = 25.0 units b) AB = 12.4 units c) AC = 31.4 units d) AE = 4.8 units e) ED = 7.6 units (This was given) f) DC = 19.0 units g) ∠ABE = 21.3◦ h) ∠BEA = 110.1◦ i) ∠BAE = 48.6◦ j) ∠BED = 69.9◦ k) ∠EDB = 65.7◦ l) ∠DBE = 44.4◦ m) ∠DBC = 43.9◦ n) ∠BDC = 114.3◦ 8. Let S1 = A, S2 = B, S3 = C. The Law of Sines may be used to determine the measure of ∠B and then either the Law of Sines or the Law of Cosines may be used to determine the length of side a. c b = sin C sin B c b = → c = 4500, ∠C = 56◦ , ∠b = 4000◦ sin C sin B 4500 4000 = → simplify sin(56◦ ) sin B 4500(sin(B)) = 4000(sin(56◦ )) → simplify 4500 sin B = 4000(0.8290) → simplify 4500 sin B = 3316.1503 → solve sin B 4500 3316.1503 → solve = 4500 4500 sin B = 0.7369 sin−1 (sin B) = sin−1 (0.7369) ∠B ≈ 47.5◦ ∠A = 180◦ − (56◦ + 47.5◦ ) ∠A = 180◦ − (103.5◦ ) ∠A = 76.5◦ 177 a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → b = 4000, c = 4500, angleA = 76.5◦ a2 = (4000)2 + (4500)2 − 2(4000)(4500) cos(76.5◦ ) → simplify √ a2 = 27845966.9 → both sides √ √ a2 = 27845966.9 → simplify a ≈ 5276.9 ft The distance between Sensor 3 and Sensor 2 is approximately 5276.9 feet. If the range of Sensor 3 is 6000 feet, it will be able to detect all movement from its location to Sensor 2. 9. Let S4 = D. ∠D = 180◦ − (36◦ + 49◦ ) ∠D = 180◦ − (85◦ ) ∠D = 95◦ The Law of Sines may be used to determine the distance between Sensor 2 and Sensor 4, as well as the distance between Sensor 3 and Sensor 4. c d = sin C sin D c d = → ∠c = 49◦ , ∠D = 95◦ , ∠d = 5276.9 sin C sin D c 5276.9 = → simplify ◦ sin(49 ) sin(95◦ ) (sin(95◦ ))c = 5276.9(sin(49◦ )) → simplify 0.9962c = 5276.9(0.7547) → simplify 0.9962c = 3982.4764 → solve 3982.4764 0.9962c = → solve 0.9962 0.9962 c ≈ 3997.7 feet The distance between Sensor 2 and Sensor 4 is approximately 3997.7 feet. 178 b d = sin B sin D b d = → ∠B = 36◦ , ∠D = 95◦ , d = 5276.9 sin B sin D b 5276.9 = → simplify ◦ sin(36 ) sin(95◦ ) (sin(95◦ ))b = 5276.9(sin(36◦ )) → simplify (0.9962)b = 5276.9(0.5875) → simplify (0.9962)b = 3101.6840 → solve 3101.6840 (0.9962)b = 0.9962 → solve 0.9962 b ≈ 3113.5 feet The distance between Sensor 3 and Sensor 4 is approximately 3113.5 feet. 10. Company A - the law of cosines may be used to determine the distance over which a driver has cell phone service. Company A a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → b = 47, c = 38, ∠A = 72.8◦ a2 = (47)2 + (38)2 − 2(47)(38) cos(72.8◦ ) → simplify √ a2 = 2596.7308 → both sides √ √ a2 = 2596.7308 → simplify a ≈ 51.0 miles Company B 179 b e = sin B sin E b e = → b = 59, e = 58, ∠B = 12◦ sin B sin E 59 58 = → simplify sin(12◦ ) sin E 59(sin E) = 58(sin(12◦ )) → simplify 59(sin E) = 58(0.2079) → simplify 59(sin E) = 12.0589 → solve 5 9(sin E) 12.0589 = → solve 59 59 sin E = 0.2044 → solve sin−1 (sin E) = sin−1 (0.2044) ∠E ≈ 11.8◦ ∠D = 180◦ − (12◦ + 11.8◦ ) ∠D = 180◦ − (23.8◦ ) ∠D = 156.2◦ b d = sin B sin D b d = → b = 59, ∠B = 12◦ , ∠D = 156.2◦ sin B sin D 59 d = → simplify ◦ sin(12 ) sin(156.2◦ ) 59(sin(156.2◦ )) = (sin(12◦ ))d → simplify 59(0.4035) = (0.2079)d → simplify 23.8092 = (0.2079)d → solve 0.2079d 23.8092 = =→ solve 0.2079 0.2079 114.5 miles ≈ d There is an overlap in cell phone service for approximately 63.5 miles. General Solutions of Triangles Review Exercises: 1. a) In the following triangle, the case AAS is given. 180 There is only one solution since the measure of two angles has been given. The Law of Sines would be used to determine the length of side b. b) In the following triangle, the case SAS is given. There is only one solution since the measure of two sides and the included angle has been given. The Law of Cosines would be used to determine the length of side c. c) In the following triangle, the case SSS is given. There is only one solution since the measure of the three sides has been given. The Law of Cosines would be used to determine the measure of ∠A. d) In the following triangle, the case SSA is given. 181 The Law of Sines would be used to determine the measure of ∠B. However, when the Law of Sines is applied, there is no solution. e) In the following triangle, the case SSA is given. There are two solutions since the measure of one angle and the length of two sides has been given. The Law of Sines would be used to determine the measure of ∠B. 2. a) a b = sin A sin B b a = → a = 22.3, ∠A = 69◦ , ∠B = 12◦ sin A sin B b 22.3 = → simplify sin(69◦ ) sin(12◦ ) 22.3(sin(12◦ )) = (sin(69◦ ))b → simplify 22.3(0.2079) = (0.9336)b → simplify 4.6362 = (0.9336)b → solve (0.9336)b 4.6362 = → solve 0.9336 0.9336 5.0 units ≈ b 182 b) c2 = a2 + b2 − 2ab cos C c2 = a2 + b2 − 2ab cos C → a = 1.4, b = 2.3, ∠C = 58◦ c2 = (1.4)2 + (2.3)2 − 2(1.4)(2.3) cos(58◦ ) → simplify √ c2 = 3.8373 → both sides √ √ c2 = 3.8373 → simplify c ≈ 2.0 units c) b2 + c2 − a2 2bc b2 + c2 − a2 cos A = → a = 3.3, b = 6.1, c = 4.8 2bc (6.1)2 + (4.8)2 − (3.3)2 cos A = → simplify 2(6.1)(4.8) 49.36 cos A = → divide 58.56 cos A = 0.8429 cos A = cos−1 (cos A) = cos−1 (0.8429) ∠A ≈ 32.6◦ d) a b = sin A sin B b a = → a = 15, b = 25, ∠A = 58◦ sin A sin B 15 25 = → simplify sin(58◦ ) sin B 15(sin B) = 25(sin(58◦ )) → simplify 15(sin B) = 25(0.8480) → simplify 15(sin B) = 21.2012 → solve 21.2012 1 5(sin B) = =→ solve 1 5 15 (sin B) = 1.4134 sin−1 (sin B) = sin−1 (1.4134) Does Not Exist 183 e) a b = sin A sin B a b = → a = 45, b = 60, ∠A = 47◦ sin A sin B 45 60 = → simplify sin(47◦ ) sin B 45(sin B) = 60(sin(47◦ )) → simplify 45(sin B) = 60(0.7314) → simplify 45(sin B) = 43.884 → simplify 4 5(sin B) 43.884 = =→ solve 4 5 45 (sin B) = 0.9752 sin−1 (sin B) = sin−1 (0.9752) 77.2◦ ≈ ∠B Or ∠B = 180◦ − 77.2◦ = 102.8◦ 3. The following information is still unknown: a) c and ∠C b) ∠A and ∠B c) ∠B and ∠C d) There is no solution e) c and ∠C 4. When solving a triangle, a check list can be used to ensure that no parts have been missed. In △ABC →a = b= ∠A = ∠B = c= ∠C = a) ∠C = 180◦ − (12◦ + 69◦ ) ∠C = 180◦ − (81◦ ) ∠C = 99◦ 184 c2 = a2 + b2 − 2ab cos C c2 = a2 + b2 − 2ab cos C → a = 22.3, b = 5.0, ∠C = 99◦ c2 = (22.3)2 + (5.0)2 − 2(22.3)(5.0) cos(99◦ ) → simplify √ c2 = 557.1749 → both sides √ √ c2 = 557.1749 → simplify c ≈ 23.6 units In △ABC →a = 22.3 ∠A = 69◦ b = 5.0 ∠B = 12◦ c = 23.6 ∠C = 99◦ SOLVED c) b2 + c2 − a2 2bc b2 + c2 − a2 cos A = → a = 1.4, b = 2.3, c = 2.0 2bc (2.3)2 + (2.0)2 − (1.4)2 → simplify cos A = 2(2.3)(2.0) 7.33 → divide cos A = 9.2 cos A = 0.7967 cos A = cos−1 (cos A) = cos−1 (0.7967) ∠A ≈ 37.2◦ ∠B = 180◦ − (58◦ + 37.2◦ ) ∠B = 180◦ − (95.2◦ ) ∠B = 84.8◦ In △ABC →a = 1.4 ∠A = 37.2◦ b = 2.3 ∠B = 84.8◦ c = 2.0 ∠C = 58◦ 185 SOLVED a2 + c2 − b2 2ac a2 + c2 − b2 cos B = → a = 3.3, b = 6.1, c = 4.8 2ac (3.3)2 + (4.8)2 − (6.1)2 cos B = → simplify 2(3.3)(4.8) −3.28 cos A = → divide 31.68 cos A = −0.1035 cos B = cos−1 (cos B) = cos−1 (−0.1035) ∠B ≈ 95.9◦ ∠A = 180◦ − (95.9◦ + 32.6◦ ) ∠A = 180◦ − (128.5◦ ) ∠A = 51.5◦ In △ABC →a = 3.3 ∠A = 51.5◦ b = 6.1 ∠B = 95.9◦ c = 2.0 ∠C = 32.6◦ SOLVED d) There is no solution. e) ∠C = 180◦ − (47◦ + 77.2◦ ) OR ∠C = 180◦ − (124.2◦ ) ∠C = 55.8◦ ∠C = 180◦ − (47◦ + 102.8◦ ) ∠C = 180◦ − (149.8◦ ) ∠C = 30.2◦ c2 = a2 + b2 − 2ab cos C c2 = a2 + b2 − 2ab cos C → a = 45, b = 60, ∠C = 55.8◦ c2 = (45)2 + (60)2 − 2(45)(60) cos(55.8◦ ) → simplify √ c2 = 2589.7498 → both sides √ √ c2 = 2589.7498 → simplify c ≈ 50.9 units c2 = a2 + b2 − 2ab cos C c2 = a2 + b2 − 2ab cos C → a = 45, b = 60, ∠C = 30.2◦ c2 = (45)2 + (60)2 − 2(45)(60) cos(30.2◦ ) → simplify √ c2 = 957.9161 → both sides √ √ c2 = 957.9161 → simplify c ≈ 31.0 units 186 In △ABC →a = 45 ∠A = 47◦ b = 60 ∠B = 77.2◦ SOLVED c = 50.9 ∠C = 55.8◦ OR In △ABC →a = 45 ∠A = 47◦ b = 60 ∠B = 103.8◦ c = 31 ∠C = 30.2◦ SOLVED 5. The area of a rhombus is readily found by using the formula A = 12 xy where x and y are the diagonals of the rhombus. These diagonals intersect at right angles. The length of the diagonal BD is 21.5 cm. and is bisected by the shorter diagonal AC. There are four right triangles within the rhombus. To determine the length of the shorter diagonal, the Pythagorean Theorem can be used. This distance can be doubled to obtain the length of AC. △BEC △BEC → BE = 1 (21.5)10.75, BC = 12(hyp) 2 In △BEC, the Pythagorean Theorem must be used to calculate the length of EC. 187 (h)2 = (s1 )2 + (s2 )2 (12)2 = (10.75)2 + (s2 )2 (12)2 = (10.75)2 + (s2 )2 √ 28.4375 = (s2 )2 √ 28.4375 = (s2 )2 5.3 cm ≈ s The length of AC is 2(5.3 cm) = 10.6 cm The area of the rhombus is: 1 xy 2 1 A = xy → x = BD(21.5 cm, y = AC(10.6 cm) 2 1 A = (21.5)(10.6) → solve 2 A ≈ 113.95 cm2 A= To calculate the measure of the angles of the rhombus, use trigonometric ratios. adj hyp 10.75 cos ∠B = 12 cos ∠B = 0.8958 opp hyp 10.75 sin ∠C = 12 sin ∠C = 0.8958 sin ∠C = cos ∠B = sin−1 (sin ∠C) = sin−1 (0.8958) ∠C ≈ 63.6◦ cos−1 (cos ∠B) = cos−1 (0.8958) ∠B ≈ 26.4◦ ∠BCD = 2(63.6◦ ) = 127.2◦ ∠ABC = 2(26.4◦ ) = 52.8◦ ∠BAD = 2(63.6◦ ) = 127.2◦ ∠ADC = 2(26.4◦ ) = 52.8◦ 6. To begin the solution to this question, begin by dividing the pentagon into 3 triangles. One triangle has vertices 1, 2, 5. The second triangle has vertices 2, 4, 5. The third triangle has vertices 2, 3, 4. 188 △125 c = a2 + b2 − 2ab cos C 2 c2 = a2 + b2 − 2ab cos C → a = 192, b = 190.5, ∠C = 81◦ c2 = (192)2 + (190.5)2 − 2(192)(190.5) cos(81◦ ) → simplify √ c2 = 61710.7560 → both sides √ √ c2 = 61710.7560 → simplify c ≈ 248.4 units a2 + c2 − b2 2ac a2 + c2 − b2 → a = 192, b = 190.5, c = 248.4 = 2ac (192)2 + (248.4)2 − (190.5)2 = → simplify 2(192)(248.4) 62276.31 = → divide 95385.6 = 0.6529 cos B = cos B cos B cos B cos B −1 cos (cos B) = cos−1 (0.6529) ∠B(∠2) ≈ 49.2◦ ∠A = 180◦ − (81◦ + 49.2◦ ) ∠A = 180◦ − (130.2◦ ) ∠A(∠5) = 49.8◦ Area of Triangle 1: 1 ab sin C 2 1 = ab sin C → a = 192, b = 190.5, ∠C = 81◦ 2 1 = (192)(190.5) sin(81◦ ) → simplify 2 1 = (192)(190.5)(0.9877) 2 = 18, 062.8 square units K= K K K K 189 △234 e = b2 + d2 − 2bd cos E 2 e2 = b2 + d2 − 2bd cos E → b = 146, d = 173.8, ∠C = 73◦ e2 = (146)2 + (173.8)2 − 2(146)(173.8) cos(73◦ ) → simplify √ e2 = 36684.6929 → both sides √ √ e2 = 36684.6929 → simplify e ≈ 191.5 units d2 + e2 − b2 2de d2 + e2 − b2 → d = 173.8, e = 191.5, b = 146 = 2de (173.8)2 + (191.5)2 − (146)2 = → simplify 2(173.8)(191.5) 45562.69 = → divide 66565.4 = 0.6845 cos B = cos B cos B cos B cos B −1 cos (cos B) = cos−1 (0.6845) ∠B(∠2) ≈ 46.8◦ ∠D = 180◦ − (73◦ + 46.8◦ ) ∠D = 180◦ − (119.8◦ ) ∠D(∠4) = 60.2◦ Area of Triangle 3: 1 bd sin E 2 1 = bd sin E → b = 146, d = 173.8, ∠E = 73◦ 2 1 = (146)(173.8) sin(73◦ ) → simplify 2 1 = (146)(173.8)(0.9563) 2 = 12, 133.0 square units K= K K K K 190 △245 a2 + d2 − b2 2ad a2 + d2 − b2 = → a = 191.5, d = 248.4, e = 118 2ad (191.5)2 + (248.4)2 − (118)2 = → simplify 2(191.5)(248.4) 84450.81 = → divide 95137.2 = 0.8877 cos B = cos B cos B cos B cos B −1 cos (cos B) = cos−1 (0.8877) ∠B(∠2) ≈ 27.4◦ b2 + d2 − a2 2bd b2 + d2 − a2 cos A = → b = 118, d = 248.4, a = 191.5 2bd (118)2 + (248.4)2 − (191.5)2 cos A = → simplify 2(118)(248.4) 38954.31 cos A = → divide 58622.4 cos A = 0.6645 cos A = cos−1 (cos A) = cos−1 (0.6645) ∠A(∠5) ≈ 48.4◦ ∠D = 180◦ − (27.4◦ + 48.4◦ ) ∠D = 180◦ − (75.8◦ ) ∠D(∠4) = 104.2◦ Area of Triangle 2: 1 ad sin B 2 1 = ad sin B → a = 191.5, d = 248.4, ∠E = 27.4◦ 2 1 = (191.5)(248.4) sin(27.4◦ ) → simplify 2 1 = (191.5)(248.4)(0.4602) 2 = 10, 945.5 square units K= K K K K Total Area: 191 18.062.8 square units + 12, 133.0 square units + 10, 945.5 square units = 41, 141.3 square units. Measure of ∠2 = 49.2◦ + 27.4◦ + 46.8◦ = 123.4◦ Measure of ∠4 = 104.2◦ + 60.2◦ = 164.4◦ Measure of ∠5 = 49.8◦ + 48.4◦ = 98.2◦ 7. This question cannot be answered. There is not enough information given. 8. If Island 4 is 22.6 miles from Island 1 and at a heading of 86.2◦ , then there is an angle of 3.8◦ made with Island 1. The distance from Island 2 to Island 4(D) c2 = d2 + b2 − 2db cos C c2 = d2 + b2 − 2db cos C → d = 28.3, b = 22.6, ∠C = 3.8◦ c2 = (28.3)2 + (22.6)2 − 2(28.3)(22.6) cos(3.8◦ ) → simplify √ c2 = 35.3023 → both sides √ √ c2 = 35.3023 → simplify c ≈ 5.9 units The distance from Island 3 to Island 4 is 52.4 miles + 5.9 miles = 58.3 miles The angle formed by Island 3 with Islands 1 and 4 192 c2 + d2 − a2 2cd c2 + d2 − a2 = → a = 22.6, c = 58.3, d = 59.8 2cd (58.3)2 + (59.8)2 − (22.6)2 = → simplify 2(58.3)(59.8) 6464.17 = → divide 6972.68 = 0.9271 cos B = cos B cos B cos B cos B −1 cos (cos B) = cos−1 (0.9271) ∠B ≈ 22.0◦ The angle formed by Island 4 with Islands 1 and 3 a2 + c2 − d2 2ac a2 + c2 − d2 = → a = 22.6, c = 58.3, d = 59.8 2ac (22.6)2 + (58.3)2 − (59.8)2 = → simplify 2(22.6)(58.3) 333.61 → divide = 2635.16 = 0.1266 cos D = cos D cos D cos D cos D −1 cos (cos D) = cos−1 (0.1266) ∠D ≈ 82.7◦ 9.a) The following diagram represents the problem. The Law of Cosines must be used to calculate the distance the ball must be shot to make it to the green in one shot. 193 b2 = a2 + c2 − 2ac cos B b2 = a2 + c2 − 2ac cos B → a = 187, c = 218, ∠B = 115◦ b2 = (187)2 + (218)2 − 2(187)(218) cos(115◦ ) → simplify √ b2 = 116949.9121 → both sides √ √ b2 = 116949.9121 → simplify b ≈ 342.0 yards b) a b = sin A sin B a b = → a = 187, b = 342, ∠B = 115◦ sin A sin B 187 342 = → simplify sin A sin(115◦ ) 187(sin(115◦ )) = 342(sin A) → simplify 187(0.9063) = 342(sin A) → simplify 169.4781 = 342(sin A) → solve 169.4781 342(sin A) = → solve 342 342 0.4959 = (sin A) sin−1 (0.4956) = sin−1 (sin A) 29.7◦ ≈ ∠A He must hit the ball within an angle of 29.7◦ . 10.a) The following diagram represents the problem. The degree of his slice is 180◦ − (162.2◦ + 14.2◦ ) = 3.6◦ 194 b) b c = sin B sin C b c = → b = 320, ∠B = 162.2◦ , ∠C = 14.2◦ sin B sin C 320 c = → simplify ◦ sin(162.2 ) sin(14.2◦ ) 320(sin(14.2◦ )) = (sin(162.2◦ ))c → simplify 320(0.2453) = (0.3057)c → simplify 78.496 = (0.3057)c → solve (0.3057)c 78.496 = → solve 0.3057 0.3057 256.8 yards ≈ c c) a c = sin A sin C a c = → c = 256.8, ∠A = 3.6◦ , ∠C = 14.2◦ sin A sin C 256.8 a = → simplify ◦ sin(3.6 ) sin(14.2◦ ) (sin(14.2◦ ))a = 256.8(sin(3.6◦ )) → simplify (0.2453)a = 256.8(0.0628) → simplify (0.2453)a = 16.1270 → solve 16.1270 (0.2453)a = =→ solve 0.2453 0.2453 a ≈ 65.7 yards Vectors Review Exercises: 1. Because m ⃗ and ⃗n are perpendicular, the Pythagorean Theorem can be used determine the magnitude of the resultant vector. To determine the direction, the trigonometric ratios can be applied. a) (h)2 = (s1 )2 + (s2 )2 (h)2 = (29.8)2 + (37.7)2 (h)2 = 2309.33 √ √ h2 = 2309.33 h ≈ 48.1 The magnitude is approximately 48.1 units. 195 opp adj 37.7 tan θ = 29.8 tan θ = 1.2651 tan θ = tan−1 (tan θ) = tan−1 (1.2651) θ ≈ 51.7◦ The direction is approximately 51.7◦ . b) (h)2 = (s1 )2 + (s2 )2 (h)2 = (29.8)2 + (5.4)2 (h)2 = 37 √ √ h2 = 37 h ≈ 6.1 The magnitude is approximately 6.1 units. 196 opp adj 5.4 tan θ = 2.8 tan θ = 1.9286 tan θ = tan−1 (tan θ) = tan−1 (1.9286) θ ≈ 62.6◦ The direction is approximately 62.6◦ . c) (h)2 = (s1 )2 + (s2 )2 (h)2 = (11.9)2 + (9.4)2 (h)2 = 229.97 √ √ h2 = 229.97 h ≈ 15.2 The magnitude is approximately 15.2 units. 197 opp adj 9.4 tan θ = 11.9 tan θ = 0.7899 tan θ = tan−1 (tan θ) = tan−1 (0.7899) θ ≈ 38.3◦ The direction is approximately 38.3◦ . d) (h)2 = (s1 )2 + (s2 )2 (h)2 = (48.3)2 + (47.6)2 (h)2 = 4598.65 √ √ (h)2 = 4598.65 h ≈ 67.8 The magnitude is approximately 67.8 units. 198 opp adj 47.6 tan θ = 48.3 tan θ = 0.9855 tan θ = tan−1 (tan θ) = tan−1 (0.9855) θ ≈ 44.6◦ The magnitude is approximately 44.6◦ . e) (h)2 = (s1 )2 + (s2 )2 (h)2 = (18.6)2 + (17.5)2 (h)2 = 652.21 √ √ (h)2 = 652.21 h ≈ 25.5 The magnitude is approximately 25.5 units. opp adj 17.5 tan θ = 18.6 tan θ = 0.9409 tan θ = tan−1 (tan θ) = tan−1 (0.9409) θ ≈ 43.3◦ The direction is approximately 43.3◦ 2. 199 Table 5.1 Operation Diagram Resultant a) ⃗a + ⃗b ⃗a + ⃗b = 6 cm + 3.2 cm = 9.2 cm b) ⃗a + d⃗ ⃗a + d⃗ = 6 cm + 4.8 cm = 10.8 cm c) ⃗c + d⃗ ⃗c + d⃗ = 1.3 cm + 4.8 cm = 6.1 cm d) ⃗a − d⃗ ⃗ = 6 cm + ⃗a − d⃗ = ⃗a + (−d) (−4.8 cm) = 1.2 cm e) ⃗b − ⃗a ⃗b−⃗a = ⃗b+(−⃗a) = 3.2 cm−6 cm = 2.8 cm 200 Table 5.1: (continued) Operation Diagram Resultant f) d⃗ − ⃗c d⃗ − ⃗c = d⃗ + (−⃗c) = 4.8 cm − 1.3 cm = 3.5 cm 3. |⃗a + ⃗b| = |⃗a| + |⃗b| is true if and only if both vectors are positive. 4. (h)2 = (s1 )2 + (s2 )2 (h)2 = (225)2 + (18)2 (h)2 = 50949 √ √ (h)2 = 50949 h ≈ 225.7 mph The plane’s speed is approximately 225.7 mph. 201 opp adj 18 tan θ = 225 tan θ = 0.08 tan θ = tan−1 (tan θ) = tan−1 (0.08) θ ≈ 4.6◦ NE The direction is approximately 4.6◦ Northeast 5. (h)2 = (s1 )2 + (s2 )2 (h)2 = (330)2 + (410)2 (h)2 = 277000 √ √ (h)2 = 277000 h ≈ 526.3 Newtons The magnitude is approximately 526.3 Newtons opp adj 410 tan θ = 330 tan θ = 1.2424 tan θ = tan−1 (tan θ) = tan−1 (1.2424) θ ≈ 51.2◦ Northeast The direction is 51.2◦ Northeast. 6. To determine the magnitude and the direction of each vector in standard position, use the coordinates of the terminal point and the coordinates of the origin in the distance formula to calculate the magnitude. 202 The x−coordinate of the terminal point represents the horizontal distance and the y−coordinate represents the vertical distance. These values can be used with the tangent function to determine the direction of the vector. a) √ (x2 − x1 )2 + (y2 − y1 )2 √ |⃗v | = (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) → (x2 , y2 ) = (12, 18) √ |⃗v | = (12 − 0)2 + (18 − 0)2 → simplify √ |⃗v | = (12)2 + (18)2 → simplify √ |⃗v | = 468 → simplify |⃗v | ≈ 21.6 |⃗v | = opp adj 18 tan θ = 12 tan θ = 1.5 tan θ = tan−1 (tan θ) = tan−1 (1.5) θ ≈ 56.3◦ b) √ (x2 − x1 )2 + (y2 − y1 )2 √ |⃗v | = (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) → (x2 , y2 ) = (−3, 6) √ |⃗v | = (−3 − 0)2 + (6 − 0)2 → simplify √ |⃗v | = (−3)2 + (6)2 → simplify √ |⃗v | = 45 → simplify |⃗v | = |⃗v | ≈ 67 opp adj 6 tan θ = −3 tan θ = −2.0 tan θ = tan−1 (tan θ) = tan−1 (2.0) θ ≈ 63.46◦ but the tangent function is negative in the 2nd quadrant. θ = 186◦ − 63.4◦ = 116.6◦ 203 c) |⃗v | = |⃗v | = |⃗v | = |⃗v | = √ √ √ √ √ (x2 − x1 )2 + (y2 − y1 )2 (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) → (x2 , y2 ) = (−1, −9) (−1 − 0)2 + (−9 − 0)2 → simplify (−1)2 + (−9)2 → simplify |⃗v | = 82 → simplify |⃗v | ≈ 9.1 opp adj −9 tan θ = −1 tan θ = 9.0 tan θ = tan−1 (tan θ) = tan−1 (9.0) θ ≈ 83.7◦ The angle is in the 3rd quadrant and has a value of 180◦ + 83.7◦ = 263.7◦ d) √ (x2 − x1 )2 + (y2 − y1 )2 √ |⃗v | = (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) → (x2 , y2 ) = (3, −2) √ |⃗v | = (3 − 0)2 + (−2 − 0)2 → simplify √ |⃗v | = (3)2 + (−2)2 → simplify √ |⃗v | = 13 → simplify |⃗v | ≈ 3.6 |⃗v | = opp adj 3 tan θ = −2 tan θ = −1.5 tan θ = tan−1 (tan θ) = tan−1 (1.5) θ ≈ 56.3◦ The angle is in the 4th quadrant and has a value of 270◦ + 56.3◦ = 326.3◦ 7. In order to determine the magnitude and direction of a vector that is not in standard position, the initial point must be translated to the origin and the terminal point translated the same number of units. For 204 example a vector with an initial point (2, 4) and a terminal point (8, 6) will become (2 − 2, 4 − 4) = (0, 0) and (8 − 2, 6 − 4) = (6, 2). Once the points of the vector in standard position have been established, the distance formula and the tangent function can be applied to determine the magnitude and the direction. a) |⃗v | = |⃗v | = |⃗v | = |⃗v | = √ √ √ √ √ (x2 − x1 )2 + (y2 − y1 )2 (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) → (x2 , y2 ) = (6, 2) (6 − 0)2 + (2 − 0)2 → simplify (6)2 + (2)2 → simplify |⃗v | = 40 → simplify |⃗v | ≈ 6.3 opp adj 2 tan θ = 6 tan θ = 0.3333 tan θ = tan−1 (tan θ) = tan−1 (0.3333) θ ≈ 18.4◦ b) √ (x2 − x1 )2 + (y2 − y1 )2 √ |⃗v | = (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) → (x2 , y2 ) = (−2, 3) √ |⃗v | = (−2 − 0)2 + (3 − 0)2 → simplify √ |⃗v | = (−2)2 + (3)2 → simplify √ |⃗v | = 13 → simplify |⃗v | = |⃗v | ≈ 3.6 opp adj 3 tan θ = −2 tan θ = −1.5 tan θ = tan−1 (tan θ) = tan−1 (1.5) θ ≈ 56.3◦ but the tangent function is negative in the 2nd quadrant θ = 180◦ − 56.3◦ = 123.7◦ 205 c) √ (x2 − x1 )2 + (y2 − y1 )2 √ |⃗v | = (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) |⃗v | = → (x2 , y2 ) = (16, −18) √ |⃗v | = (16 − 0)2 + (−18 − 0)2 → simplify √ |⃗v | = (16)2 + (−18)2 → simplify √ |⃗v | = 580 → simplify |⃗v | ≈ 24.1 opp adj −18 tan θ = 16 tan θ = −1.125 tan θ = tan−1 (tan θ) = tan−1 (1.125) θ ≈ 48.4◦ The angle is in the 4th quadrant and has a value of 360◦ − 48.4◦ = 311.6◦ d) √ (x2 − x1 )2 + (y2 − y1 )2 √ |⃗v | = (x2 − x1 )2 + (y2 − y1 )2 → (x1 , y1 ) = (0, 0) |⃗v | = → (x2 , y2 ) = (10, 10) √ |⃗v | = (10 − 0)2 + (10 − 0)2 → simplify √ |⃗v | = (10)2 + (10)2 → simplify √ |⃗v | = 200 → simplify |⃗v | ≈ 14.1 opp adj 10 tan θ = 10 tan θ = 1.0 tan θ = tan−1 (tan θ) = tan−1 (1.0) θ ≈ 45◦ 8. To determine the magnitude of the resultant vector and the angle it makes with a, the parallelogram method will have to be used. The opposite angles of a parallelogram are congruent as are the opposite sides. The Law of Cosines can be used to calculate the magnitude of the resultant vector. 206 a) If ∠CDA = 65◦ then ∠ABC = 65◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(65◦ ) + 2∠BCD = 360◦ 130◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 130◦ 2∠BCD = 230◦ ◦ 2∠BCD = 230 2 2 ∠BCD = 115◦ In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 10, d = 13, ∠A = 115◦ a2 = (10)2 + (13)2 − 2(10)(13) cos(115◦ ) → simplify √ a2 = 378.8807 → both sides √ √ a2 = 378.8807 → simplify a ≈ 19.5 units 207 sin A sin D = a d sin A sin D = → ∠A = 115◦ , a = 19.5, ∠d = 13 a d sin(115◦ ) sin D = → simplify 19.5 13 ◦ sin(115 )(13) = (19.5) sin D → simplify (0.9063)(13) = (19.5)(sin D) → simplify 11.7820 = (19.5) sin D → solve sin D (19.5) 11.7820 = → solve 19.5 19.5 0.6042 = sin D → solve sin−1 (0.6042) = sin−1 (sin D) 37.2◦ ≈ ∠D b) If ∠CDA = 119◦ then ∠ABC = 119◦ ∠BCD = ∠BAD 208 ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(119◦ ) + 2∠BCD = 360◦ 238◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 238◦ 2∠BCD = 122◦ ◦ 2∠BCD = 122 2 2 ∠BCD = 61◦ In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 32, d = 25, ∠A = 61◦ a2 = (32)2 + (25)2 − 2(32)(25) cos(61◦ ) → simplify √ a2 = 873.3046 → both sides √ √ a2 = 873.3046 → simplify a ≈ 29.6 units sin A sin D = a d sin A sin D = → ∠A = 61◦ , a = 29.6, ∠d = 32 a d sin(61◦ ) sin D = → simplify 29.6 32 sin(61◦ )(32) = (29.6) sin D → simplify (0.8746)(32) = (29.6)(sin D) → simplify 27.9872 = (29.6) sin D → solve sin D (29.6) 27.9872 = → solve 29.6 29.6 0.9455 = sin D → solve sin −1 (0.9455) = sin−1 (sin D) 71.0◦ ≈ ∠D 209 c) If ∠CDA = 132◦ then ∠ABC = 132◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(132◦ ) + 2∠BCD = 360◦ 264◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 264◦ 2∠BCD = 96◦ 2∠BCD 96◦ = 2 2 ∠BCD = 48◦ In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 31, d = 31, ∠A = 48◦ a2 = (31)2 + (31)2 − 2(31)(31) cos(48◦ ) → simplify √ a2 = 635.9310 → both sides √ √ a2 = 635.9310 → simplify a ≈ 25.2 units 210 sin A sin D = a d sin A sin D = → ∠A = 48◦ , a = 25.2, d = 31 a d sin D sin(48◦ ) = → simplify 25.2 31 sin(48◦ )(31) = (25.5) sin D → simplify (0.7431)(31) = (25.2)(sin D) → simplify 23.0361 = (25.2) sin D → solve sin D 23.0361 (25.2) = → solve 25.2 25.2 0.9141 = sin D → solve sin−1 (0.9141) = sin−1 (sin D) 66.1◦ ≈ ∠D d) If ∠CDA = 26◦ then ∠ABC = 26◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(26◦ ) + 2∠BCD = 360◦ 52◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 52◦ 2∠BCD = 308◦ 2∠BCD 308◦ = 2 2 ∠BCD = 154◦ In △BAD 211 a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 29, d = 44, ∠A = 154◦ a2 = (29)2 + (44)2 − 2(29)(44) cos(15.4◦ ) → simplify √ a2 = 5070.7224 → both sides √ √ a2 = 5070.7224 → simplify a ≈ 71.2 units sin A sin D = a d sin A sin D = → ∠A = 154◦ , a = 71.2, d = 29 a d sin D sin(154◦ ) = → simplify 71.2 29 sin(154◦ )(29) = (71.2) sin D → simplify (0.4384)(29) = (71.2)(sin D) → simplify 12.7136 = (71.2) sin D → solve sin D 12.7136 (71.2) = → solve 71.2 71.2 0.1786 = sin D → solve sin−1 (0.1786) = sin−1 (sin D) 10.3◦ ≈ ∠D 9. To solve this problem, it must be noted that the angle of 48◦ is made with the horizontal and is located outside of the parallelogram. The angle inside of the parallelogram is the diﬀerence between the angle made with the horizontal by car A and the angle made with the horizontal by car B. This angle is 39◦ . The solution may now be completed by using the Law of Cosines to determine the magnitude and the Law of Sines to calculate the direction of the resultant. 212 If ∠CDA = 39◦ then ∠ABC = 39◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(39◦ ) + 2∠BCD = 360◦ 78◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 78◦ 2∠BCD = 282◦ 2∠BCD 282◦ = 2 2 ∠BCD = 141◦ In △BAD 213 a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 35, d = 52, ∠A = 141◦ a2 = (35)2 + (52)2 − 2(35)(52) cos(141◦ ) → simplify √ a2 = 6757.8113 → both sides √ √ a2 = 6757.8113 → simplify a ≈ 82.2 units sin D sin A = a d sin A sin D = → ∠A = 141◦ , a = 82.2, d = 52 a d sin(141◦ ) sin D = → simplify 82.2 52 sin(141◦ )(52) = (82.2) sin D → simplify (0.6293)(52) = (82.2)(sin D) → simplify 32.7236 = (82.2) sin D → solve sin D 32.7236 (82.2) = → solve 82.2 82.2 0.3981 = sin D → solve sin−1 (0.3981) = sin−1 (sin D) 23.5◦ ≈ ∠D The direction is this result plus the angle that car A makes with the horizontal. 23.5◦ + 48◦ = 71.5◦ 10. To solve this problem, the Law of Cosines must be used to determine the magnitude of the resultant and the Law of Sines to calculate the direction that the resultant makes with the smaller force. 214 If ∠CDA = 25.4◦ then ∠ABC = 25.4◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(25.4◦ ) + 2∠BCD = 360◦ 50.8◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 50.8◦ 2∠BCD = 309.2◦ 2∠BCD 309.2◦ = 2 2 ∠BCD = 154.6◦ In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 3750, d = 4210, ∠A = 154.6◦ a2 = (3750)2 + (4210)2 − 2(3750)(4210) cos(154.6◦ ) → simplify √ a2 = 60309411.87 → both sides √ √ a2 = 60309411.87 → simplify a ≈ 7, 765.9 ≈ 7, 766 lbs. sin A sin D = a d sin D sin A = → ∠A = 154.6◦ , a = 7766, d = 3750 a d sin(154.6◦ ) sin D = → simplify 7766 3750 sin(154.6◦ )(3750) = (7766) sin D → simplify (0.4289)(3750) = (7766)(sin D) → simplify 1608.375 = (7766) sin D → solve sin D 1608.375 (7766) = → solve 7766 7766 0.2071 = sin D → solve sin−1 (0.2071) = sin−1 (sin D) 12◦ ≈ ∠D Component Vectors Review Exercises: 215 1. To determine the resulting ordered pair, simply apply scalar multiplication. a) ⃗a = 2⃗b ⃗ → ⃗b = (0, 0) to (5, 4) ⃗a = 2b ⃗ → 2(5, 4) = (10, 8) ⃗a = 2b ⃗a = (10, 8) ⃗a = (0, 0) to (10, 8) b) 1 ⃗a = − ⃗c 2 1 ⃗a = − ⃗c → c = (0, 0) to (−3, 7) 2 1 1 ⃗a = − ⃗c → c = − (−3, 7) = (1.5, −3.5) 2 2 ⃗a = (1.5, −3.5) ⃗a = (0, 0) to (1.5, −3.5) c) ⃗a = 0.6⃗b ⃗a = 0.6⃗b → ⃗b = (0, 0) to (5, 4) ⃗a = 0.6⃗b → ⃗b = 0.6(5, 4) = (3, 2.4) ⃗a = (3, 2.4) ⃗a = (0, 0) to (3, 2.4) d) ⃗a = −3⃗b ⃗a = −3⃗b → ⃗b = (0, 0) to (5, 4) ⃗a = −3⃗b → b = −3(5, 4) = (−15, −12) ⃗a = (−15, −12) ⃗a = (0, 0) to (−15, −12) 2. To determine the magnitude of the vertical and horizontal components of these vectors, add the absolute values of the coordinates necessary to return the initial point to the origin with the absolute value of the coordinates of the terminal point. a) horizontal = |3| + |2| = 5 vertical = | − 8| + | − 1| = 9 b) horizontal = | − 7| + |11| = 18 vertical = | − 13| + |19| = 32 c) horizontal = | − 4.2| + | − 1.3| = 5.5 vertical = |6.8| + | − 9.4| = 16.2 d) horizontal = | − 5.23| + | − 0.237| = 5.467 vertical = | − 4.98| + |0| = 4.98 216 3. To determine the magnitude of the horizontal and vertical components if the resultant vector’s magnitude and direction are given, use the trigonometric ratio for cosine to determine the magnitude of the horizontal component and the trigonometric ratio for sine to determine the magnitude of the vertical component. When calculating these values, consider the direction to be an angle in standard position and the magnitude of the resultant to be the hypotenuse ⃗q of the right triangle ⃗q⃗r⃗s. a) |⃗r| r = |⃗q| q r cos 35◦ = 75 r 0.8192 = 75 ( ) r 75(0.8192) = 7 5 7 5 |61.4| ≈ r(horizontal) 61.4 ≈ r(horizontal) |⃗s| s = |⃗q| q s sin 35◦ = 75 s 0.5736 = 75 (s) 75(0.5736) = 75 75 |43| ≈ s(vertical) 43 ≈ s(vertical) r |⃗r| = |⃗q| q r cos 162◦ = 3.4 r −0.9511 = 3.4 ( ) r 3.4(−0.9511) = 3.4 3.4 | − 3.2| ≈ r(horizontal) 3.2 ≈ r(horizontal) |⃗s| s = |⃗q| q s sin 162◦ = 3.4 s 0.3090 = 3.4 ) ( s 3.4(0.3090) = 3.4 3.4 |1.1| ≈ s(vertical) 1.1 ≈ s(vertical) cos 35◦ = sin 35◦ = b) sin 162◦ = cos 162◦ = c) |⃗s| s = |⃗q| q s sin 12◦ = 15.9 s 0.2079 = 15.9 ( ) s 15.9(0.2079) = 15.9 15.9 r |⃗r| = |⃗q| q r cos 12◦ = 15.9 r 0.9781 = 15.9 ( ) r 15.9(0.9781) = 15.9 15.9 |15.6| ≈ r(horizontal) 15.6 ≈ r(horizontal) sin 12◦ = cos 12◦ = |3.3| ≈ s(vertical) 3.3 ≈ s(vertical) 217 d) |⃗s| s = |⃗q| q s sin 223◦ = 189.27 s − 0.6820 = 189.27 r |⃗r| = |⃗q| q r cos 223◦ = 189.27 r −0.7314 = 189.27 ( 189.27(−0.7314) = 189.27 sin 223◦ = cos 223◦ = r ) 189.27 | − 138.4| ≈ r(horizontal) 189.27(−0.6820) = 189.27 | − 129.1| ≈ s(vertical) 138.4 ≈ r(horizontal) ( s ) 189.27 129.1 ≈ s(vertical) 4. To determine the magnitude and the direction of the resultant vector, the Pythagorean Theorem can be use to calculate the magnitude and the trigonometric ratio for sine can be used to determine the angle that it makes with the smaller force. (⃗q)2 = (⃗r)2 + (⃗s)2 −−→ −−→ (⃗q)2 = (32.1)2 + (8.50)2 (⃗q)2 = 1102.66 √ √ (⃗q)2 = 1102.66 ⃗q ≈ 33.2 Newtons |⃗s| s = |⃗q| q 8.50 sin x = 33.2 sin x = 0.2560 sin x = sin−1 (sin x) = sin−1 (0.2560) x ≈ 14.8◦ 5. To determine the magnitude of the resultant and the angle it makes with the larger force, the parallelogram method must be used. Once the diagram has been sketched, the Law of Cosines and the Law of Sines can be used. If ∠CDA = 43◦ then ∠ABC = 43◦ 218 ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(43◦ ) + 2∠BCD = 360◦ 86◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 86◦ 2∠BCD = 274◦ ◦ 2∠BCD = 274 2 2 ∠BCD = 137◦ In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 140, d = 186, ∠A = 137◦ a2 = (140)2 + (186)2 − 2(140)(186) cos(137◦ ) → simplify √ a2 = 92284.9008 → both sides √ √ a2 = 92284.9008 → simplify a ≈ 303.8 ≈ 304 Newtons sin A sin D = a d sin A sin D = → ∠A = 137◦ , a = 304, d = 186 a d sin D sin(137◦ ) = → simplify 304 186 sin(137◦ )(186) = (304) sin D → simplify (0.6820)(186) = (304)(sin D) → simplify 126.852 = (304) sin D → solve sin D (304) 126.852 = → solve 304 304 0.4173 = sin D → solve sin−1 (0.4173) = sin−1 (sin D) 24.7◦ ≈ ∠D This angle is counterclockwise from the smaller force. 6. To determine the magnitude of the resultant and the angle it makes with ⃗a, the parallelogram method must be used. Once the diagram has been sketched, the Law of Cosines and the Law of Sines can be used. 219 a) If ∠CDA = 144◦ then ∠ABC = 144◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(144◦ ) + 2∠BCD = 360◦ 288◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 288◦ 2∠BCD = 72◦ 2∠BCD 72◦ = 2 2 ∠BCD = 36◦ In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 22, d = 49, ∠A = 36◦ a2 = (22)2 + (49)2 − 2(22)(49) cos(36◦ ) → simplify √ a2 = 1140.7594 → both sides √ √ a2 = 1140.7594 → simplify a ≈ 33.8 units 220 sin A sin D = a d sin A sin D = → ∠A = 36◦ , a = 33.8, d = 22 a d sin(36◦ ) sin D = → simplify 33.8 22 sin(36◦ )(22) = (33.8) sin D → simplify (0.5878)(22) = (33.8)(sin D) → simplify 12.9316 = (33.8) sin D → solve sin D 12.9316 (33.8) → solve = 33.8 33.8 0.3826 = sin D → solve sin−1 (0.3826) = sin−1 (sin D) 22.5◦ ≈ ∠D This angle is from the horizontal. b) If ∠CDA = 28◦ then ∠ABC = 28◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(28◦ ) + 2∠BCD = 360◦ 56◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 56◦ 2∠BCD = 304◦ ◦ 2∠BCD = 304 2 2 ∠BCD = 152◦ In △BAD 221 a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 19, d = 71, ∠A = 152◦ a2 = (19)2 + (71)2 − 2(19)(71) cos(152◦ ) → simplify √ a2 = 7784.1926 → both sides √ √ a2 = 7784.1926 → simplify a ≈ 88.2 units sin A sin D = a d sin A sin D = → ∠A = 152◦ , a = 88.2, d = 19 a d sin(152◦ ) sin D = → simplify 88.2 19 sin(152◦ )(19) = (88.2) sin D → simplify (0.4695)(19) = (88.2) sin D → simplify (8.9205) = (88.2)(sin D) → solve sin D 8.9205 (88.2) = → solve 88.2 88.2 0.1011 = sin D → solve sin−1 (0.1011) = sin−1 (sin D) 5.8◦ ≈ ∠D This angle is from the horizontal. 222 c) If ∠CDA = 81◦ then ∠ABC = 81◦ ∠BCD = ∠BAD ∠CDA + ∠ABC + ∠BCD + ∠BAD = 360◦ 2∠CDA + 2∠BCD = 360◦ 2(81◦ ) + 2∠BCD = 360◦ 162◦ + 2∠BCD = 360◦ 2∠BCD = 360◦ − 162◦ 2∠BCD = 198◦ ◦ 2∠BCD = 198 2 2 ∠BCD = 99◦ 223 In △BAD a2 = b2 + d2 − 2bd cos A a2 = b2 + d2 − 2bd cos A → b = 5.2, d = 12.9, ∠A = 99◦ a2 = (5.2)2 + (12.9)2 − 2(5.2)(12.9) cos(99◦ ) → simplify √ a2 = 214.4372 → both sides √ √ a2 = 214.4372 → simplify a ≈ 14.6 units sin A sin D = a d sin A sin D = → ∠A = 99◦ , a = 14.6, d = 5.2 a d sin(99◦ ) sin D = → simplify 14.6 5.2 ◦ sin(99 )(5.2) = (14.6) sin D → simplify (0.9877)(5.2) = (14.6)(sin D) → simplify 5.1360 = (14.6) sin D → solve sin D 5.1360 (14.6) → solve = 14.6 14.6 0.3518 = sin D → solve sin−1 (0.3518) = sin−1 (sin D) 20.6◦ ≈ ∠D This angle is from the horizontal. 7. To determine the horizontal and vertical components, use the trigonometric ratio for cosine to calculate the horizontal component and the ratio for sine to calculate the vertical component. 224 |⃗r| r = |⃗q| q r ◦ cos 28.2 = 12 r 0.8813 = 12 ( ) r 12(0.8813) = 1 2 1 2 |10.6| ≈ r(horizontal) 10.6 ≈ r(horizontal) |⃗s| s = |⃗q| q s ◦ sin 28.2 = 12 s 0.4726 = 12 (s) 12(0.4726) = 12 12 |5.67| ≈ s(vertical) 5.67 ≈ s(vertical) cos 28.2◦ = sin 28.2◦ = 8. To determine the heading of the plane, the Law of Cosines must be used to determine the magnitude of the plane and then the Law of Sines to calculate the heading. 225 a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → ∠A = 118◦ , b = 42, c = 155 a2 = (42)2 + (155)2 − 2(42)(155) cos(118◦ ) → simplify √ a2 = 31901.5198 → both sides √ √ a2 = 31901.5198 → simplify a ≈ 178.6 km/h 226 sin A sin B = a b sin A sin B = → ∠A = 118◦ , a = 178.6, b = 42 a b sin(118◦ ) sin B = → simplify 178.6 42 sin(118◦ )(42) = (178.6) sin B → simplify (0.8829)(42) = (178.6)(sin B) → simplify 37.0818 = (178.6) sin B → solve sin B 37.0818 (178.6) = → solve 178.6 178.6 0.2076 = sin B → solve sin−1 (0.2076) = sin−1 (sin B) 12◦ ≈ ∠B The heading is 12◦ + 83◦ ≈ 95◦ 9. The ﬁrst step is to apply the Pythagorean to determine the speed the boat will travel with the current to cross the river and then use the trigonometric ratio for sine to calculate the angle at which the boat must travel. (h)2 = (s1 )2 + (s2 )2 (h)2 = (10)2 + (2)2 (h)2 = 104 √ √ h2 = 104 10.2 mph ≈ h opp hyp opp sin A = → opp = 2.00, hyp = 10.2 hyp 2.00 sin A = → simplify 10.2 sin A = 0.1961 sin A = sin−1 (sin A) = sin−1 (0.1961) ∠A ≈ 11.3◦ 227 10. If AB is any vector, then BA is a vector of the same magnitude but in the opposite direction. AB + (−BA) = (0, 0). Real-World Triangle Problem Solving Review Exercises: 1. To determine the distance from the command post to a point on the ground directly below the helicopter, use the trigonometric ratio for tangent. opp adj opp tan A = → opp = 2500, adj = x, ∠A = 9.3◦ adj 2500 tan(9.3◦ ) = → simplify x 2500 (0.1638) = → simplify x( ) 2500 (0.1638)(x) = (x) → simplify x tan A = (0.1638)(x) = 2500 → solve (0.1638)(x) 2500 = → solve .1638 .1638 x ≈ 15262.5 feet 2. To determine the distance across the canyon, use the trigonometric ratio for tangent. 228 opp adj opp → opp = 387.6, adj = x, ∠B = 67◦ tan B = adj 387.6 tan(67◦ ) = → simplify x 387.6 (2.3559) = → simplify x( ) 387.6 → simplify (2.3559)(x) = (x) x tan B = (2.3559)(x) = 387.6 → solve (2.3559)(x) 387.6 = 2.3559 → solve 2.3559 x ≈ 164.5 feet 3. To determine the distance between the stoplights on Street A, use the Trigonometric ratio for Sine. 229 opp hyp opp sin A = → opp = x, hyp = 0.5, ∠A = 54◦ hyp x sin(54◦ ) = → simplify 0.5 x → simplify (0.8090) = 0.5 ( ) x → solve (0.8090)(0.5) = (0.5) 0.5 0.4 miles ≈ x sin A = 4. To determine the distance that the ball was shot and the distance of the second baseman from the ball, the Law of Sines and/or the Law of Cosines may be used. 230 ∠C = 180◦ − (127◦ + 18◦ ) ∠C = 180◦ − (145◦ ) ∠C = 35◦ Distance the ball was hit c b = sin C sin B b c = → c = 127.3, ∠C = 35◦ , ∠B = 127◦ sin C sin B 127.3 b = → simplify ◦ sin(35 ) sin(127◦ ) 127.3 b = → simplify 0.5734 0.7986 127.3(0.7986) = (0.5734)b → simplify 101.6663 = (0.5734)b → solve 101.6663 (0.5734)b = → solve 0.5734 0.5734 177.2 feet ≈ b Distance the ball is from the second baseman c a = sin C sin A c a = → c = 127.3, ∠C = 35◦ , ∠A = 18◦ sin C sin A 127.3 a = → simplify sin(35◦ ) sin(18◦ ) a 127.3 = → simplify 0.5734 0.3090 127.3(0.3090) = (0.5734)a → simplify 39.3379 = (0.5734)a → solve 39.3379 (0.5734)b = → solve 0.5734 0.5734 68.6 feet ≈ a 5. There is not enough information given in this question to answer it. 6. To determine the distance from the Tower to Target 2, the Law of Cosines must be used. Target 1 = A Target 2 = B Tower = C 231 a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → ∠A = 67.2◦ , b = 18, c = 37 a2 = (18)2 + (37)2 − 2(18)(37) cos(67.2◦ ) → simplify √ a2 = 1176.8292 → both sides √ √ a2 = 1176.8292 a ≈ 34.3 miles The sensor will not be able to detect the second target. Target 2 is out of range by approximately 4.3 miles. 7. To determine the number of bacteria, the area of the lake must be calculated. The Law of Cosines must be used to determine the measure of one of the angles of the triangle. Then the formula K = 21 bc sin A can be used to calculate the area of the lake. Dock 1 = A Dock 2 = B Dock 3 = C b2 + c2 − a2 2bc b2 + c2 − a2 cos A = → a = 587, b = 396, c = 247 2bc (396)2 + (247)2 − (587)2 cos A = → simplify 2(396)(247) −126744 cos A = → divide 195624 cos A = −0.6479 cos A = cos−1 (cos A) = cos−1 (−0.6479) ∠A ≈ 130.4◦ Area of lake: 1 bc sin A 2 1 K = bc sin A → b = 396, c = 247, ∠A = 130.4◦ 2 1 K = (396)(247) sin(130.4◦ ) → simplify 2 K = 37243.8 ft2 K= The numbers of bacteria that are living on the surface of the lake are 37243.8(5.2 × 1013 ) ≈ 1.94 × 1018 8. A direction of 37◦ east of north is an angle of 53◦ with the horizontal. This must be considered when drawing the diagram to represent the problem and when calculating the distance from Tower B to the ﬁre. This distance can be calculated by using the Law of Cosines. 232 a2 = b2 + c2 − 2bc cos A a2 = b2 + c2 − 2bc cos A → ∠A = 53◦ , b = 45, c = 100 a2 = (45)2 + (100)2 − 2(45)(100) cos(53◦ ) → simplify. √ a2 = 6608.6648 → both sides √ √ a2 = 6608.6648 a ≈ 81.3 miles 9. The two forces are acting at right angles to each other due to the direction of the forces. The Pythagorean Theorem can be used to determine the magnitude of the resultant on the footing and the tangent function may be used to calculate the direction of the resultant. a) (h)2 = (s1 )2 + (s2 )2 (h)2 = (1870)2 + (2075)2 (h)2 = 7802525 √ √ h2 = 7802525 2793.3 lbs. ≈ h 233 b) opp adj opp tan C = → opp = 2075, adj = 1870 adj tan C = 1.1096 → simplify tan C = tan−1 (tan C) = tan−1 (1.1096) ∠C ≈ 48◦ 10. A heading of 118◦ is an angle of 62◦ with the horizontal. A second heading of 34◦ will result in an angle of 62◦ + 34◦ = 96◦ . b2 = a2 + c2 − 2ac cos B b2 = a2 + c2 − 2ac cos C → ∠C = 96◦ , a = 215, c = 342 b2 = (215)2 + (342)2 − 2(215)(342) cos(96◦ ) → simplify. √ b2 = 178560.9558 → both sides √ √ b2 = 178560.9558 b ≈ 422.6 km 234 sin C sin B = b c sin B sin C = → ∠B = 96◦ , b = 422.6, c = 342 b c sin(96◦ ) sin C = → simplify 422.6 342 sin(96◦ )(342) = (422.6) sin C → simplify (0.9945)(342) = (422.6) sin C → simplify 340.119 = (422.6) sin C → solve sin D (422.6) 340.119 → solve = 422.6 422.6 0.8048 = sin C → solve sin−1 (0.8048) = sin−1 (sin C) 53.6◦ ≈ ∠C The heading is 53.6◦ + 34◦ ≈ 87.6◦ 235 236 Chapter 6 TE Polar Equations and Complex Numbers - Solution Key 6.1 Polar Equations and Complex Numbers Polar Coordinates Review Exercises 1. To plot these points using computer software, choose polar as the grid. Then enter the coordinates. a) 237 b) ) ( 2. To determine four pair of polar coordinates to represent the point A −4, π4 , use the formula (r, θ + 2πk) and choose diﬀerent values for k. Then use the formula (r, θ + [2k + 1]π) and again choose diﬀerent values for k. 1 2 Using (r, θ + 2πk) and k = −1 Using (r, θ + 2πk) and k = − π (r, θ + 2πk) → r = −4, θ = , k = −1 4 ( ) π −4, + 2π(−1) → simplify 4 ( ) π −4, − 2π → common deno min ator 4 ( ) π 8π −4, − → simplify 4 4 ( ) 7π −4, − 4 π 1 (r, θ + 2πk) → r = −4, θ = , k = − 4 2 ( ( )) π 1 −4, + 2π − → simplify 4 2 ( ) π −4, + (−1)π → simplify 4 ( ) π −4, − π → common deno min ator 4 ( ) π 4π −4, − → simplify 4 4 ( ) 3π −4, − 4 238 Using (r, θ + [2k + 1]π) and k = −1 π (r, θ + [2k + 1]π) → r = 4, θ = , k = −1 4 ( π ) 4, + [2(−1) + 1]π → simplify 4 ( π ) 4, + [−2 + 1]π → simplify 4 ( π ) 4, + [−2 + 1]π → simplify 4 ) ( π 4, + [−1]π → simplify 4 ) ( π 4, − π → common deno min ator 4 ( ) π 4π 4, − → simplify 4 4 ( ) 3π 4, − 4 Using (r, θ + [2k + 1]π) and k = 0 π (r, θ + [2k + 1]π) → r = 4, θ = , k = 0 4 ( π ) 4, + [2(0) + 1]π → simplify 4 ( π ) 4, + [0 + 1]π → simplify 4 ( π ) 4, + π → common deno min ator ( 4 ) π 4π 4, + → simplify 4 4 ( ) 5π 4, 4 ( ) First Pair → −4, − 7π 4 ( ) Second Pair → −4, − 3π 4 ( ) Third Pair → 4, − 3π 4 ( ) Fourth Pair → 4, 5π 4 3. To calculate the distance between the points use the√distance formula for polar coordinates which is a form of)the Law Use the formula p1 p2 = r12 + r22 − 2r1 r2 cos(θ2 − θ1 ) and the coordinates ( ( of Cosines. ) r 1 , θ1 r2 , θ2 and . 1, 30◦ 6, 135◦ √ p1 p2 = p1 p2 = p1 p2 = √ √ r12 + r22 − 2r1 r2 cos(θ2 − θ1 ) r12 + r22 − 2r1 r2 cos(θ2 − θ1 ) → r1 = 1, r2 = 6, θ1 = 30◦ , θ2 = 135◦ (1)21 + (6)2 − 2(1)(6) cos(135◦ − 30◦ ) → simplify √ p1 p2 = 40.1058 → simplify p1 p2 ≈ 6.33 units Sinusoids of one Revolution (e.g. limaçons, cardioids) Review Exercises 1. 239 a) A limaçon with an inner loop b) A cardioid 240 c) A dimpled limaçon 2. For the equation r = 4 cos 2θ such that 0◦ ≤ θ ≤ 360◦ , create a table of values and sketch the graph. Repeat the process for r = 4 cos 3θ such that 0◦ ≤ θ ≤ 360◦ θ 4 cos 2θ 0◦ 4 30◦ 2 60◦ −2 90◦ −4 120◦ −2 150◦ 2 180◦ 4 210◦ 2 240◦ −2 270◦ −4 300◦ −2 330◦ 2 360◦ 4 θ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦ 4 cos 3θ 4 0 −4 0 4 0 −4 0 4 0 −4 0 4 The number n has an aﬀect on the number of petals on the rose. The ﬁrst graph, r = 4 cos 2θ such that 0◦ ≤ θ ≤ 360◦ the rose has four petals on it. In this case, n is an even, positive integer and the rose has an 241 even number of petals. The second graph, r = 4 cos 3θ such that 0◦ ≤ θ ≤ 360◦ the rose has three petals on it. In this case, n is an odd, positive integer and the rose has an odd number of petals. Graphs of Polar Equations Review Exercises 1. To determine the rectangular coordinates of polar coordinates means to express the given point as (x, y). To do this use the formula x = r cos θ to determine the x−coordinate and the formula y = r sin θ to determine the y−coordinate. ( ) a) A −4, 5π 4 x = r cos θ y = r sin θ 5π x = r cos θ → r = −4, θ = 4 ( ) 5π x = (−4) cos → simplify 4 √ ( ) 2 5π 5π x = −4 cos → cos =− 4 4 2 ( √ ) 2 x = −4 − → simplify 2 ( √ ) 2 x = −2 4 − → solve 2 √ x=2 2 ( ) √ √ 5π A −4, = (2 2, 2 2) 4 5π y = r sin θ → r = −4, θ = 4 ( ) 5π y = (−4) sin → simplify 4 √ ( ) 2 5π 5π y = −4 sin → sin =− 4 4 2 ( √ ) 2 y = −4 − → simplify 2 ( √ ) 2 y = −24 − → solve 2 √ y=2 2 b) B(−3, 135◦ ) x = r cos θ → r = −3, θ = 135◦ x = (−3) cos(135◦ ) → simplify √ 2 x = −3 cos 135 → cos 135 = − 2 ( √ ) 2 → simplify x = −3 − 2 √ 3 2 x= (2 √ √ ) 3 2 3 2 ◦ B(−3, 135 ) = , 2 2 ◦ ◦ ( ) c) C 5, 2π 3 242 y = r sin θ → r = −3, θ = 135◦ y = (−3) sin(135◦ ) → simplify √ 2 y = −3 sin(135 ) → sin 135 = − 2 (√ ) 2 y = −3 → simplify 2 √ 3 2 y=− 2 ◦ ◦ x = r cos θ y = r sin θ 2π x = r cos θ → r = 5, θ = 3 ( ) 2π x = (5) cos → simplify 3 ( ) 2π 1 2π x = (5) cos → cos =− 3 3 2 ( ) 1 x = (5) − → simplify 2 2π y = r sin θ → r = 5, θ = 3 ( ) 2π y = (5) sin → simplify 3 √ ( ) 2π 2π 3 y = (5) sin → sin = 3 3 2 (√ ) 3 → simplify y=5 2 √ 5 3 y= 2 5 → solve 2 x = −2.5 √ ) ( ) ( 2π 5 3 C 5, = −2.5, 3 2 x=− 2. r = 6 cos θ The following graph represents a circle with its center at (3, 0) and a radius of 3 units. r = 6 cos θ r2 = 6 cos θ x2 + y 2 = 6 cos θ → letx = cos θ x2 + y 2 = 6x → simplify x2 + y 2 − 6x = 6x − 6x → simplify (x2 − 6x) + y 2 = 0 → complete the square (x2 − 6x + 9) + y 2 = 0 + 9 → write as a perfect square trinomial (x − 3)2 + y 2 = 9 (x − h)2 + (y − k)2 = r2 → general formula (x − 3)2 + (y − 0)2 = 32 Rectangular to Polar Review Exercises 243 √ 1. To write rectangular coordinates in polar form, use the formula r = x2 + y 2 to determine the value of r and the formula θ = Arc tan xy + π for x < 0 or the formula θ = Arc tan xy for x > 0 to calculate the value of θ. a) A(−2, 5). This point is located in the 2nd quadrant and x < 0. √ x2 + y 2 √ r = x2 + y 2 → x = −2, y = 5 y + πfor x < 0 x y = Arc tan + π → x = −2, y = 5 x 5 = Arc tan + π → simplify −2 = tan−1 (−2.5) + π → simplify = −1.1903 + π → simplify ≈ 1.95 θ = Arc tan r= θ √ (−2)2 + (5)2 → simplify √ r = 29 → simplify r ≈ 5.39 r= θ θ θ θ A(−2, 5) = (5.39, 1.95) b) B(5, −4). This point is located in the 4th quadrant and x > 0. √ x2 + y 2 √ r = x2 + y 2 → x = 5, y = −4 y θ = Arc tan for x < 0 x y θ = Arc tan → x = 5, y = −4 x −4 θ = Arc tan → simplify 5 θ = tan−1 (−0.8) → simplify r= √ (5)2 + (−4)2 → simplify √ r = 41 → simplify r= r ≈ 6.40 B(5, −4) = (6.40, −0.67) θ = −0.67 2. To write the equation (x − 4)2 + (y − 3)2 = 25, expand the equation in terms of x and y. Then replace x with the expression r cos θ and y with r sin θ. (x − 4)2 + (y − 3)2 = 25 (x − 4)2 + (y − 3)2 = 25 → exp and (x2 − 8x + 16) + (y 2 − 6y + 9) = 25 → simplify x2 − 8x + y 2 − 6y + 25 = 25 → simplify x2 − 8x + y 2 − 6y + 25 − 25 = 25 − 25 → simplify x2 − 8x + y 2 − 6y = 0 → x = r cos θ, y = r sin θ, r = x2 + y 2 r2 − 8(r cos θ) − 6(r sin θ) = 0 → simplify r2 − 8r cos θ − 6r sin θ = 0 → common factor r(r − 8 cos θ − 6 sin θ) = 0 → solve r = 0 or r − 8 cos θ − 6 sin θ = 0 The graph of r = 0 is a single point – the origin. The graph of r − 8 cos θ − 6 sin θ = 0 contains this single point. The polar form of (x − 4)2 + (y − 3)2 = 25 as a single equation is r = 8 cos θ + 6 sin θ and the graph is 244 The graph was drawn on a polar grid and then the grid was deleted so as to reveal a clear view of the shape of the graph - a circle with its center at (4, 3) and a radius of 5 units. The circumference of the circle passes through the origin. Polar Equations and Complex Numbers Review Exercises 1. To prove that the equation represents a parabola, write the equation in standard form. Then determine the vertex (h, k), the focus (h + p, k) and the directrix (x = h − p). y 2 − 4y − 8x + 20 = 0 y 2 − 4y − 8x + 20 + 8x − 20 = 8x − 20 → simplify y 2 − 4y = 8x − 20 → complete the square y 2 − 4y + 4 = 8x − 20 + 4 → simplify y 2 − 4y + 4 = 8x − 16 → perfect square binomial (y − 2)2 = 8x − 16 → common factor (y − 2)2 = 8(x − 2) The equation is in standard form. The vertex (h, k) is (2, 2). 8 4p 4p = 8 → = → p = 2. 4 4 Therefore the focus is (h + p, k) which equals (2 + 2, 2) → (4, 2). The directrix, (x = h − p) is (x = 2 − 2) → x = 0. 2. To determine the center (h, k), the vertices (h ± a, k), foci and the eccentricity the equation in standard form. 245 (c) a of the ellipse, express 9x2 + 16y 2 + 54x − 32y − 47 = 0 9x2 + 16y 2 + 54x − 32y − 47 + 47 = 0 + 47 → simplify 9x2 + 16y 2 + 54x − 32y = 47 → common factor 9x2 + 54x + 16y 2 − 32y = 47 → common factor 9(x2 + 6x) + 16(y 2 − 2y) = 47 → complete the square 9(x2 + 6x + 9) + 16(y 2 − 2y + 1) = 47 → add to right side 9(x2 + 6x + 9) + 16(y 2 − 2y + 1) = 47 + 81 + 16 → simplify 9(x2 + 6x + 9) + 16(y 2 − 2y + 1) = 144 → perfect square binomial 9(x + 3)2 + 16(y − 1)2 = 144 → ÷(144) 16(y − 1)2 9(x + 3)2 + 144 144 2 9 (x + 3) 1 6(y − 1)2 + 144(16) 144(9) (x + 3)2 (y − 1)2 + 2 4 33 2 (x − h) (y − k)2 + a2 b2 = 144 → simplify 144 = 1 → simplify =1 = 1 → s tan dard form The centre is (h, k) → (−3, 1). The vertices are (h ± a, k) and a = 4. Thus the vertices are (−3 ± 4, 1) → (1, 1) and (−7, 1). √ √ √ √ √ The foci are (h ± a, k) and c = a2 − b2 → c = 42 − 32 → 16 − 9 = 7 The foci are (−3 ± 7, 1) ≈ (−5.65, 1) and (−0.35, 1). ( ) √ The eccentricity ac is 47 ≈ 0.66. 3. To determine the eccentricity, the type of conic and the directrix, use the general formula r = de 1−e cos θ . de 1 − e cos θ 2 r= → ÷(4) 4 − cos θ r= r= r= 4 4 − 2 4 1 4 cos θ → simplify 0.5 1 − 0.25 cos θ If 0 < e < 1, the graph will be an ellipse. The eccentricity is 0.25 so the conic is an ellipse. The numerator 0.5 de = 0.5. Therefore the directrix is de e → 0.25 = 2. The directrix is x = −2. Graph and Calculate Intersections of Polar Curves Review Exercises 246 1. To determine the points of intersection of the graphs, hide the grid when the graph has been completed. This makes it easier to determine the intersection. Then, solve the equations for each graph. a) r = sin(3θ) and r = 3 sin θ There appears to be one point of intersection – the origin. Let r = 0 r sin(3θ) 0 = sin 3θ r = 3 sin θ 0 = 3 sin θ 0 3 sin θ = 3 3 0 = sin θ sin−1 (0) = sin−1 (sin θ) sin−1 (0) = sin−1 (sin θ) sin−1 (0) = sin−1 (sin θ) 0=θ 0=θ To accommodate 3θ, multiplying does not change the value of θ. The point of intersection is (0, 0) → r = 0, θ = 0 b) Plot the graphs of r = 2 + 2 sin θ and r = 2 − 2 cos θ. 247 There appears to be three points of intersection. One point of intersection seems to be the origin (0, 0). Let θ = 0 r = 2 + 2 sin θ r = 2 + 2 sin θ → θ = 0 r = 2 − 2 cos θ r = 2 − 2 cos θ → θ = 0 r = 2 + 2 sin(0) → sin 0 = 0 r = 2 + 2(0)simplify r = 2 + 2 cos(0) → cos 0 = 0 r = 2 − 2(1)simplify r =2+0 r=2 r =2−2 r=0 The coordinates represent the same point (0, 0). r = 2 + 2 sin θ r = 2 − 2 cos θ 248 2 + 2 sin θ = 2 − 2 cos θ 2 − 2 + 2 sin 2 sin 2 sin 2 cos sin cos tan θ θ θ θ θ θ θ = 2 − 2 cos θ → simplify = −2 cos θ → ÷(2 cos θ) −2 cos θ = → ÷simplify 2 cos θ sin θ = −1 → = tan θ cos θ = −1 tan−1 (tan θ) = tan−1 (1) π θ= 4 The tangent function is negative in the 2nd and 4th quadrants. 2nd Quadrant π θ=π− 4 π θ = π − → common deno min ator 4 4π π θ= − → simplify 4 4 3π θ= 4 4th Quadrant π θ = 2π − 4 π θ = 2π − common deno min ator 4 8π π θ= − → simplify 4 4 7π θ= 4 r = 2 + 2 sin θ r = 2 − 2 cos θ 3π r = 2 + 2 sin θ → θ = 4 ( ) 3π r = 2 + 2 sin → simplify 4 ( ) ( ) 3π 3π r = 2 + 2 sin → sin = 0.7071 4 4 3π r = 2 − 2 cos θ → θ = 4 ( ) 3π r = 2 − 2 cos → simplify 4 ( ) ( ) 3π 3π r = 2 − 2 cos → cos = (0.7071) 4 4 r = 2 + 2(0.7071) → simplify r ≈ 3.41 7π r = 2 + 2 sin θ → θ = ( ) 4 7π r = 2 + 2 sin θ → simplify 4 ( ) ( ) 7π 7π r = 2 + 2 sin θ → sin θ = −0.7071 4 4 r = 2 − 2(0.7071) → simplify r ≈ 3.41 7π r = 2 − 2 cos θ → θ = ( ) 4 7π r = 2 − 2 cos θ → simplify 4 ( ) ( ) 7π 7π r = 2 − 2 cos θ → cos θ = −0.7071 4 4 r = 2 + 2(−0.7071) → simplify r = 2 − 2(−0.7071) → simplify r ≈ 0.59 r ≈ 0.59 Substituting the points into the equation r = 2 − 2 cos θ is not necessary but it does conﬁrm the points. ( ) ( ) 7π The points of intersection are 3.41, 3π and (0, 0). 4 , 0.59, 4 249 Equivalent Polar Curves Review Exercises 1. To write the equation in polar form, use the formulas r2 = x2 + y 2 and x = r cos θ. x2 + y 2 = 6x x2 + y 2 = 6x → x2 + y 2 , x = r cos θ r2 = 6(r cos θ) → simplify r2 = 6r cos θ → ÷(r) 6r cos θ r2 (r) = → simplify r r r = 6 cos θ 250 Both equations r = cos θ and x2 + y 2 = 6x produced the same graph – a circle with center (3, 0) and a radius of 3. ( ) ( ) 2. If the equations r = 7 − 3 cos π3 and r = 7 − 3 cos − π3 produce the same graph, then the equations are equivalent. 251 Yes, the both equations are equivalent. They are graphed above on separate axes but both could be plotted on the same grid. Only one graph would appear. Recognize Review Exercises √ √ √ Recognize i = −1, −x = i x 1. To express the square root of a negative number in terms of i, express the radicand as the product of a positive number and (−1). Then write this product as the product of the square root of the positive factor and the square root of (−1). a) √ −64 √ √ (64)( −1) = 8i b) − √ −108 √ − (108)(−1) √ √ (− 108)( −1) √ √ (− (36)(3))( −1) √ = −6i 3 252 c) √ ( −15)2 √ √ (( 15)( −1))2 √ (i 15)2 = 15i2 → i2 = −1 (−1)(15) = −15 d) √ √ ( −49)( −25) √ √ ( (49)(−1))( (25)(−1)) √ √ √ √ ( 49)( −1)( 25)( −1) (7)(i)(5)(i) = 35i2 → i2 = −1 (−1)(35) = −35 Standard Forms of Complex Numbers C Review Exercises 1. To simplify each complex number means to write it in standard form (a + bi). The conjugate is of the form (a + bi) with the same ‘a’ but the opposite ‘bi’. Example: The conjugate of 4 − 3i is 4 + 3i. a) − √ √ −400 √ − 1 − (400)(−1) √ √ √ − 1 − ( (400))( −1) → simplify = −1 − 20i √ 1− The conjugate is − 1 + 20i b) √ √ −36 √ (36)(−1) + (36)(−1) √ √ √ 36 + ( 36)( −1) → simplify √ −36i2 + 6 + 6i2 The conjugate is 6 − 6i 253 2. Solve the equation for the variables x and y. 6i − 7 = x − yi 6i − 7 − 6i + 7 = 3 − x − yi − 6i + 7 → simplify 0 = 10 − x − iy − 6i → simplify 0 + x + yi = 10 − x − yi − 6i + x + yi → simplify x + yi = 10 − 6i → solve x = 10andyi = −6i → solve x = 10 yi = −6i → ÷(i) y i −6i = i i y = −6 The values of x = 10 and y = −6 satisfy the equation 6i − 7 = 3 − x − yi. The Set of Complex Numbers (complex, real, irrational, rational, etc) Review Exercises: None – Simply an information lesson Complex Number Plane Review Exercises 1. The absolute value of a complex number in standard form (a + bi) is the square root of a2 + b2 . In other √ words |a + bi| = a2 + b2 . a) The coordinates of the points plotted on the complex number plane are: A(−5 − 3i) B(6 + 2i) |6 + 2i| = C(2 − 5i) |2 − 5i| = √ √ 40 ≈ 6.3 29 ≈ 5.4 √ | − 2 + 4i| = 20 ≈ 4.5 D(−2 + 4i) E(3 + 6i) The absolute values of the other 3 points are shown above. The detailed solutions for A and E are shown below. The question requires only two points to be done. 254 A(−5 − 3i) A(−5 − 3i) → a = −5, b = −3 √ |a + bi| = a2 + b2 √ | − 5 − 3i| = a2 + b2 → (−5 − 3i) → a = −5, b = −3 √ | − 5 − 3i| = (−5)2 + (−3)2 → simplify √ | − 5 − 3i| = 25 + 9 → simplify √ | − 5 − 3i| = 34 → simplify √ | − 5 − 3i| = 34 ≈ 5.8 E(3 + 6i) E(3 + 6i) →→ a = 3, b = 6 √ |a + bi| = a2 + b2 √ |3 + 6i| = a2 + b2 → (3 + 6i) → a = 3, b = 6 √ |3 + 6i| = (3)2 + (6)2 → simplify √ |3 + 6i| = 9 + 36 → simplify √ |3 + 6i| = 45 → simplify √ |3 + 6i| = 45 ≈ 6.7 Quadratic Formula Review Exercises 1. To describe the nature of the roots, the value of the discriminant must be calculated. If the value of the discriminant b2 + 4ac is less than zero, the roots will be a complex conjugate pair of roots. Set the equation equal to zero. 5x2 − x + 5 = 6x + 1 5x2 − x + 5 − 6x − 1 = 6x + −6x − 1 → simplify 5x2 − 7x + 4 = 0 → simplify 5x2 − 7x + 4 = 0 → a = 5, b = −7, c = 4 b2 − 4ac b2 − 4ac = (−7)2 − 4(5)(4) → evaluate b2 − 4ac = −31 b2 − 4ac < 0 → a complex conjugate pair of roots Solve the equation using the quadratic formula: x = √ −b± b2 −4ac 2a 255 5x2 − 7x + 4 = 0 → a = 5, b = −7, c = 4 √ −b ± b2 − 4ac x= 2a √ −(−7) ± (−7)2 − 4(5)(4) x= → simplify 2(5) √ 7 ± 49 − 80 x= → simplify √10 7 ± −31 → simplify x= 10 √ √ 7 ± ( 31)( −1) x= → solve √ 10 7 + i 31 x= → evalute 10 7 + 5.6i x= → evalute 10 7 + 5.6i x= ≈ 0.7 + 0.56i 10 √ 7 − i 31 → evalute 10 7 − 5.6i x= → evalute 10 7 − 5.6i x= ≈ 0.7 − 0.56i 10 x= 2. The above parabola does not intersect the x−axis. This means that the value of the discriminant, b2 − 4ac, will be less than zero. The roots of his quadratic function will be a complex, conjugate pair. Sums and Diﬀerences of Complex Numbers Review Exercises 256 1. The steps involved in adding and subtracting real numbers also apply to complex numbers. To subtract is actually adding the opposite and adding involves following the rules for integers. a) Graphically Subtract the complex numbers: (7 − 3i) − (8 − 7i) (7 − 3i) − (8 − 7i) → add(−8 + 7i) (7 − 3i) − (−8 + 7i) → simplify (7 − 8) + (−3i + 7i) → simplify = −1 + 4i Check: (7 − 3i) − (8 − 7i) (7 − 3i) − (−8 + 7i) (7 − 8) + (−3i + 7)i = −1 + 4i b) Graphically: 257 Add the complex numbers: (4.5 − 2.0i) + (6.0 + 8.5i) (4.5 + 6.0) + (−2.0i + 8.5i) → simplify (4.5 + 6.0) + (−2.0 + 8.5)i → simplify = 10.5 + 6.5i Check: (4.5 − 2.0i) + (6.0 + 8.5i) (4.5 + 6.0) + (−2.0 + 8.5)i = 10.5 + 6.5i Products and Quotients of Complex Numbers (conjugates) Review Exercises 1. To perform the operation of multiplication in part ‘a’, apply the distributive property and simplify the answer. In part ‘b’, multiply the numerator and the denominator of the fraction by the conjugate of the denominator. Apply the distributive property and simplify the answer. a) 258 (7 − 5i)(4 − 9i) (7 − 5i)(4 − 9i) → expand 7(4 − 9i) − 5i(4 − 9i) → distributive property 28 − 63i − 20i + 45i2 → simplify 28 − 83i + 45i2 → i2 = −1 28 − 83i − 45(−1) → simplify 28 − 83i − 45 → simplify = −17 − 83i b) 4 + 7i 9 − 5i 4 + 7i → multiply by(9 + 5i) 9 − 5i ( )( ) 4 + 7i 9 + 5i → simplify 9 − 5i 9 + 5i (4 + 7i)(9 + 5i) → exp and (9 − 5i)(9 + 5i) 4(9 + 5i) + 7i(9 + 5i) distributive property 9(9 + 5i) − 5i(9 + 5i) 36 + 20i + 63i + 35i2 → simplify 81 + 45i − 45i − 25i2 36 + 83i + 35i2 → i2 = −1 81 − 25i2 36 + 83i + 35(−1) → simplify 81 − 25(−1) 36 + 83i − 35 → simplify 81 + 25 1 + 83i → simplify 106 1 + 83i ≈ 0.009 + 0.783i 106 The Trigonometric or Polar Form of a Complex Number r cis θ Review Exercises 1.To express the point (6 − 8i) graphically, plot the point as you would the point (6, −8). The y−axis is the Imaginary axis and the x−axis is the Real axis. To write (6, −8) in its polar form, the value of ‘r’ must be determined as well the measure of theta. In addition,x = r cos θ and y = r sin θ. 259 6 − 8i x = 6 and y = −8 √ r = x2 + y 2 → det er min ethe value ofr √ r = (6)2 + (−8)2 → simplify √ r = 100 → simplify r = 10 opp y = adj x −8 tan θ = → divide 6 tan θ = −1.3333 tan θ = tan−1 (tan θ) = tan−1 (1.3333) θ = 53.1◦ The tangent function is negative in the 4th quadrant and the point 6 − 8i is located there. The measure of θ is 360◦ − 53.1◦ = 306.9◦ In polar form 6 − 8i is 10(cos 306.9◦ + i sin 306.9◦ ) or 10∠306.9◦ . 2. 260 ( π π) 3 cos + i sin 4 4 r=3 √ 2 π x = cos = 4 √2 π 2 y = sin = 4 2 (√ √ ) ( π π) 2 2 3 cos + i sin is actually3 +i 4 4 2 2 √ √ 3 2 3 2 + i = 2 2 De Moivre’s Theorem Review Exercises 1. The ﬁrst step in solving this problem is to express the equation in polar form. 261 √ 1 3 z =− +i 2 2 (√ ) √ 1 3 3 1 z =− +i x = − ,y = 2 2 2 2 √ r = x2 + y 2 v u( )2 ( √ ) u 1 3 t → simplify r= − + 2 2 √ 1 3 r= + → simplify 4 4 √ r= 1=1 opp y = adj x √ 3 tan θ = − → simplify 1 √ tan θ = − 3 √ −1 tan (tan θ) = tan−1 ( 3) tan θ = θ ≈ 60◦ The point is located in the 2nd quadrant and the tangent function is negative here. The measure of theta is 180◦ − 60◦ = 120◦ . The polar form of z = − 21 + i √ 3 2 is z = 1(cos 120◦ + i sin 120◦ ). Apply De Moivre’s Theorem z n = [r(cos θ + i sin θ)]n = rn (cos θ + i sin n θ) z 3 = 13 [cos 3(120◦ ) + i sin 3(120◦ )] z 3 = 13 (cos 360◦ + i sin 360◦ ) z 3 = 1(1 + i(0)) z3 = 1 2. To write the expression [2(cos 315◦ + i sin 315◦ )]3 in rectangular form, simply work backwards and apply De Moivre’s Theorem 262 [2(cos 315◦ + i sin 315◦ )]3 7π 4 n n z = [r(cos θ + i sin θ)] = rn (cos θ + i sin n θ) ( ( ) ( )) 7π 7π z n = 23 cos 3 + i sin 3 4 4 ( ) 21π 21π 21π z 3 = 8 cos + i sin (3rd quadrant) → 4 4 4 [2(cos 315◦ + i sin 315◦ )]3 → r = 2 and θ = 315◦ → → Both are negative √ ) 2 2 z3 = 8 − −i → simplify 2 2 ( √ √ ) 2 2 3 z = 8(4) − −i → simplify 2 2 √ √ z 3 = −4 2 − 4i 2 ( √ nth Root Theorem Review Exercises 1. To determine the cube root of 27i, write it as a complex number, calculate the value of r and the measure of θ √ 3 √ 3 27i 27i → (a + bi) 1 (0 + 27i) 3 → a = 0 and b = 27 → x = 0 and y = 27 Calculate the value of ‘r’: √ x2 + y 2 √ r = (0)2 + (27)2 → simplify r = 27 r= 263 π 2 [ ( √ π π )] 13 3 27i = 27 cos + i sin → simplify [ ( 2 ( ) 2 ( ) )] √ √ 1 π 1 π 3 3 27i = 27 cos + i sin → simplify 3 2 3 2 √ ( √ π π) π 3 π 1 3 27i = 3 cos + i sin → cos = , sin = 6 2 6 2 ) 6 (√ 6 √ 1 3 3 +i → simplify 27i = 3 2 2 ) (√ √ 3 1 3 + i 27i = 3 2 2 θ= 2. To determine the principal root means to calculate the positive root. The square root of a number can be ± The principal root is the positive root only. 1 (1 + i) 5 1 (a + bi) 5 1 (1 + i) 5 → a = 1, b = 1 → x =, y = 1 Calculate the value of ‘r’: √ x2 + y 2 √ r = (1)2 + (1)2 → simplify √ r= 2 r= opp y = adj x 1 tan θ = → simplify 1 tan θ = 1 tan θ = tan−1 (tan θ) = tan−1 (1) √ 2 π = θ≈ 2 4 [√ ( 1 π )] 15 π (1 + i) 5 = 2 cos + i sin → simplify (4 ( ) 4 [ ( ) )] √ 1 1 1 π 1 π + i sin → simplify (1 + i) 5 = ( 2) 5 cos 5 4 5 4 ( √ 1 π) π 5 (1 + i) 5 = 2 cos + i sin → evaluate 20 20 264 polar From : (√ 2, π) 4 1 1 (1 + i) 5 = (1.07 + 1.07i) → s tan dard from. This is the principal root of (1 + i) 5 . Solve Equations Review Exercises: 1. To solve the equation x4 + 1 = 0, an expression for determining the fourth roots of the equation, must be written. Calculate the value of ‘r’ and the measure of θ. x4 + 1 = 0 x4 + 1 − 1 = 0 − 1 → solve x4 = −1 x4 = −1 + 0i x4 = −1 + 0i → x = −1, y = 0 Calculate the value of ‘r’: √ x2 + y 2 √ r = x2 + y 2 → x = −1, y = 0 √ r = (−1)2 + (0)2 → simplify √ r= 1=1 r= opp y = adj x 0 tan θ = → simplify −1 ) ( 0 +π tan θ = −1 ( ) 0 −1 −1 tan (tan θ) = tan +π −1 tan θ = Polar Form: (1, π) θ=π 265 1 1 (−1 + 0i) 4 = [1(cos(π + 2πk)) + i sin(π + 2πk)] 4 ( ) 1 1 π + 2πk π + 2πk (−1 + 0i) 4 = (1) 4 cos + i sin →k 4 4 ( ) π π x1 = 1 cos + i sin →k=0 4 4 ( ) 1 1 π + 2πk π + 2πk 4 4 (−1 + 0i) = (1) cos + i sin →k 4 4 ( ) 3π 3π x2 = 1 cos + i sin →k=1 4 4 ( ) 1 1 π + 2πk π + 2πk + i sin (−1 + 0i) 4 = (1) 4 cos →k 4 4 ( ) 5π 5π x3 = 1 cos + i sin →k=2 4 4 ( ) 1 1 π + 2πk π + 2πk + i sin (−1 + 0i) 4 = (1) 4 cos →k 4 4 ( ) 7π 7π + i sin x4 = 1 cos →k=3 4 4 √ √ ( π) 2 2 π → +i x1 = 1 cos + i sin 4 4 2 √ 2 √ ( ) 3π 2 2 3π x2 = 1 cos + i sin →− +i 4 4 2 2 √ √ ( ) 5π 2 2 5π x3 = 1 cos + i sin →− −i 4 4 2 2 √ √ ( ) 2 2 7π 7π x4 = 1 cos + i sin → −i 4 4 2 2 266 =0 =1 =2 =3

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertisement