Instructor’s Solutions Manual, Section 5.1 Exercise 1 Solutions to Exercises, Section 5.1 1 1. Find all numbers t such that ( 3 , t) is a point on the unit circle. 1 solution For ( 3 , t) to be a point on the unit circle means that the sum of the squares of the coordinates equals 1. In other words, 1 2 3 + t 2 = 1. √ 8 8 This simpliﬁes to the equation t 2 = 9 , which implies that t = 3 or √ √ √ √ √ √ 8 t = − 3 . Because 8 = 4 · 2 = 4 · 2 = 2 2, we can rewrite this as t= √ 2 2 3 √ 2 or t = − 2 3 . Student Solution Manual, Section 5.1 Exercise 3 3. Find all numbers t such that (t, − 25 ) is a point on the unit circle. 2 solution For (t, − 5 ) to be a point on the unit circle means that the sum of the squares of the coordinates equals 1. In other words, 2 2 t 2 + − 5 = 1. This simpliﬁes to the equation t 2 = t= √ 21 − 5 . 21 25 , which implies that t = √ 21 5 or Instructor’s Solutions Manual, Section 5.1 Exercise 5 5. Find the points where the line through the origin with slope 3 intersects the unit circle. solution The line through the origin with slope 3 is characterized by the equation y = 3x. Substituting this value for y into the equation for the unit circle (x 2 + y 2 = 1) gives x 2 + (3x)2 = 1, √ 10 √ 10 which simpliﬁes to the equation 10x 2 = 1. Thus x = 10 or x = − 10 . Using each of these values√of x along with the equation y = 3x gives the √ √10 3√10 10 3 10 points 10 , 10 and − 10 , − 10 as the points of intersection of the line y = 3x and the unit circle. Student Solution Manual, Section 5.1 Exercise 7 7. Suppose an ant walks counterclockwise on the unit circle from the point (1, 0) to the endpoint of the radius that forms an angle of 70◦ with the positive horizontal axis. How far has the ant walked? solution We need to ﬁnd the length of the circular arc on the unit circle 70π 7π corresponding to a 70◦ angle. This length equals 180 , which equals 18 . Instructor’s Solutions Manual, Section 5.1 Exercise 9 9. What angle corresponds to a circular arc on the unit circle with length π 5? solution Let θ be such that the angle of θ degrees corresponds to an π θπ π arc on the unit circle with length 5 . Thus 180 = 5 . Solving this equation for θ, we get θ = 36. Thus the angle in question is 36◦ . Student Solution Manual, Section 5.1 11. Exercise 11 What angle corresponds to a circular arc on the unit circle with length 5 2? solution Let θ be such that the angle of θ degrees corresponds to an arc θπ on the unit circle with length 52 . Thus 180 = 52 . Solving this equation for θ, we get θ = 450 π . Thus the angle in question is equal to 143.2◦ . 450 ◦ π , which is approximately Instructor’s Solutions Manual, Section 5.1 Exercise 13 13. Find the lengths of√ both circular arcs on the unit circle connecting the 2 √2 points (1, 0) and 2 , 2 . √ √ solution The radius of the unit circle ending at the point 22 , 22 horizontal axis. One of the cirmakes an angle of 45◦ with the positive √2 √2 cular arcs connecting (1, 0) and 2 , 2 is shown below as the thickened √2 √2 circular arc; the other circular arc connecting (1, 0) and 2 , 2 is the unthickened part of the unit circle below. 45π π The length of the thickened arc below is 180 , which equals 4 . The entire unit circle has length 2π . Thus the length of the other circular arc below π 7π is 2π − 4 , which equals 4 . 45 1 The thickened circular arc has length 7π The other circular arc has length 4 . π 4. Student Solution Manual, Section 5.1 Exercise 15 For each of the angles in Exercises 15–20, ﬁnd the endpoint of the radius of the unit circle that makes the given angle with the positive horizontal axis. 15. 120◦ solution The radius making a 120◦ angle with the positive horizontal axis is shown below. The angle from this radius to the negative horizontal axis equals 180◦ − 120◦ , which equals 60◦ as shown in the ﬁgure below. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 60◦ ; thus the other angle must be 30◦ , as labeled below: 120 30 60 1 The side of the right triangle opposite the 30◦ angle has length √ 3 2 . 1 2; the Looking side of the right triangle opposite the 60◦ angle has length at the ﬁgure above, we see that the ﬁrst coordinate of the endpoint of the Instructor’s Solutions Manual, Section 5.1 Exercise 15 radius is the negative of the length of the side opposite the 30◦ angle, and the second coordinate of the endpoint of the radius is the length of the√ side opposite the 60◦ angle. Thus the endpoint of the radius is 1 3 −2, 2 . Student Solution Manual, Section 5.1 Exercise 17 17. −30◦ solution The radius making a −30◦ angle with the positive horizontal axis is shown below. Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 30◦ ; thus the other angle must be 60◦ , as labeled below. The side of the right triangle opposite the 30◦ angle has length √ 3 2 . ◦ 1 2; the Looking side of the right triangle opposite the 60 angle has length at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is the length of the side opposite the 60◦ angle, and the second coordinate of the endpoint of the radius is the negative of the length of the side opposite the 30◦ angle. Thus the endpoint of the radius is √3 1 2 , −2 . 30 1 60 Instructor’s Solutions Manual, Section 5.1 Exercise 19 19. 390◦ solution The radius making a 390◦ angle with the positive horizontal axis is obtained by starting at the horizontal axis, making one complete counterclockwise rotation, and then continuing for another 30◦ . The resulting radius is shown below. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 30◦ ; thus the other angle must be 60◦ , as labeled below. The side of the right triangle opposite the 30◦ angle has length 12 ; the side √ 3 opposite the 60◦ angle has length 2 . Looking at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is the length of the side opposite the 60◦ angle, and the second coordinate of the endpoint of the radius is the length √of the side opposite the 30◦ angle. Thus the 3 1 endpoint of the radius is 2 , 2 . 60 30 1 Student Solution Manual, Section 5.1 Exercise 21 For Exercises 21–26, ﬁnd the angle the radius of the unit circle ending at the given point makes with the positive horizontal axis. Among the inﬁnitely many possible correct solutions, choose the one with the smallest absolute value. 1 √3 21. − 2 , 2 √ solution Draw the radius whose endpoint is − 12 , 23 . Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle. The hypotenuse of this right triangle is a radius of the unit circle and thus has length 1. The horizontal side has √ 1 3 length 2 and the vertical side of this triangle has length 2 because the 1 √3 endpoint of the radius is − 2 , 2 . 120 30 60 1 Thus we have a 30◦ - 60◦ - 90◦ triangle, with the 30◦ angle opposite the 1 horizontal side of length 2 , as labeled above. Because 180 − 60 = 120, Instructor’s Solutions Manual, Section 5.1 Exercise 21 the radius makes a 120◦ angle with the positive horizontal axis, as shown above. In addition to making a 120◦ angle with the positive horizontal axis, this radius also makes with the positive horizontal axis angles of 480◦ , 840◦ , and so on. This radius also makes with the positive horizontal axis angles of −240◦ , −600◦ , and so on. But of all the possible choices for this angle, the one with the smallest absolute value is 120◦ . Student Solution Manual, Section 5.1 23. √2 2 ,− Exercise 23 √ 2 2 √ √ solution Draw the radius whose endpoint is 22 , − 22 . Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle. The hypotenuse of this right triangle is a radius of √ the unit circle and thus has length 1. The horizontal √ side has 2 2 length 2 and the vertical side of this triangle also has length 2 because √ √2 2 the endpoint of the radius is 2 , − 2 . 45 1 45 Thus we have here an isosceles right triangle, with two angles of 45◦ as labeled above. In addition to making a −45◦ angle with the positive horizontal axis, this radius also makes with the positive horizontal axis angles of 315◦ , 675◦ , and so on. This radius also makes with the positive horizontal axis angles of −405◦ , −765◦ , and so on. But of all the possible choices for this angle, the one with the smallest absolute value is −45◦ . Instructor’s Solutions Manual, Section 5.1 25. − Exercise 25 √ √ 2 2 2 ,− 2 √ √ solution Draw the radius whose endpoint is − 22 , − 22 . Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle. The hypotenuse of this right triangle is a radius of √ the unit circle and thus has length 1. The horizontal √ side has 2 2 length 2 and the vertical side of this triangle also has length 2 because √ √2 2 the endpoint of the radius is − 2 , − 2 . 45 45 1 135 Thus we have here an isosceles right triangle, with two angles of 45◦ as labeled above. Because the radius makes a 45◦ angle with the negative horizontal axis, it makes a −135◦ angle with the positive horizontal axis, as shown below (because 135◦ = 180◦ − 45◦ ). In addition to making a −135◦ angle with the positive horizontal axis, this radius also makes with the positive horizontal axis angles of 225◦ , 585◦ , and so on. This radius also makes with the positive horizontal axis Student Solution Manual, Section 5.1 Exercise 25 angles of −495◦ , −855◦ , and so on. But of all the possible choices for this angle, the one with the smallest absolute value is −135◦ . Instructor’s Solutions Manual, Section 5.1 Exercise 27 27. Find the lengths of both circular arcs on the unit circle connecting the 1 √3 point 2 , 2 and the point that makes an angle of 130◦ with the positive horizontal axis. 1 √3 solution The radius of the unit circle ending at the point 2 , 2 makes with the positive horizontal axis. One of the circular arcs an angle of 60◦ √ 1 3 connecting 2 , 2 and the point that makes an angle of 130◦ with the positive horizontal axis is shown below as the thickened circular arc; the other circular arc connecting these two points is the unthickened part of the unit circle below. The thickened arc below corresponds to an angle of 70◦ (because 70◦ = 70π 130◦ − 60◦ ). Thus the length of the thickened arc below is 180 , which 7π equals 18 . The entire unit circle has length 2π . Thus the length of the 7π 29π other circular arc below is 2π − 18 , which equals 18 . 60 130 7π 1 The thickened circular arc has length 18 . 29π The other circular arc has length 18 . Student Solution Manual, Section 5.1 Exercise 29 29. Find the √ lengths √ of both circular arcs on the unit circle connecting the 2 2 point − 2 , − 2 and the point that makes an angle of 125◦ with the positive horizontal axis. √ √2 2 solution The radius of the unit circle ending at − 2 , − 2 makes an axis 225◦ = 180◦ +45◦ ). angle of 225◦ with the positive horizontal √ (because √2 2 One of the circular arcs connecting − 2 , − 2 and the point that makes an angle of 125◦ with the positive horizontal axis is shown below as the thickened circular arc; the other circular arc connecting these two points is the unthickened part of the unit circle below. The thickened arc below corresponds to an angle of 100◦ (because 100◦ = 100π 225◦ − 125◦ ). Thus the length of the thickened arc below is 180 , which 5π equals 9 . The entire unit circle has length 2π . Thus the length of the 5π 13π other circular arc below is 2π − 9 , which equals 9 . 125 225 1 The thickened circular arc has length 13π The other circular arc has length 9 . 5π 9 . Instructor’s Solutions Manual, Section 5.1 Exercise 31 31. What is the slope of the radius of the unit circle that has a 30◦ angle with the positive horizontal axis? solution The radius of the unit circle that has a 30◦ angle with the positive horizontal axis has its initial point at (0, 0) and its endpoint at √3 1 , . Thus the slope of this radius is 2 2 1 −0 √2 3 2 −0 which equals √1 , 3 which equals √ 3 3 . , Student Solution Manual, Section 5.2 Exercise 1 Solutions to Exercises, Section 5.2 In Exercises 1–8, convert each angle to radians. 1. 15◦ solution Start with the equation 360◦ = 2π radians. Divide both sides by 360 to obtain 1◦ = π radians. 180 Now multiply both sides by 15, obtaining 15◦ = π 15π radians = radians. 180 12 Instructor’s Solutions Manual, Section 5.2 3. −45◦ solution Start with the equation 360◦ = 2π radians. Divide both sides by 360 to obtain 1◦ = π radians. 180 Now multiply both sides by −45, obtaining −45◦ = − π 45π radians = − radians. 4 180 Exercise 3 Student Solution Manual, Section 5.2 Exercise 5 5. 270◦ solution Start with the equation 360◦ = 2π radians. Divide both sides by 360 to obtain 1◦ = π radians. 180 Now multiply both sides by 270, obtaining 270◦ = 3π 270π radians. radians = 2 180 Instructor’s Solutions Manual, Section 5.2 7. 1080◦ solution Start with the equation 360◦ = 2π radians. Divide both sides by 360 to obtain 1◦ = π radians. 180 Now multiply both sides by 1080, obtaining 1080◦ = 1080π radians = 6π radians. 180 Exercise 7 Student Solution Manual, Section 5.2 In Exercises 9–16, convert each angle to degrees. 9. 4π radians solution Start with the equation 2π radians = 360◦ . Multiply both sides by 2, obtaining 4π radians = 2 · 360◦ = 720◦ . Exercise 9 Instructor’s Solutions Manual, Section 5.2 11. π 9 radians solution Start with the equation 2π radians = 360◦ . Divide both sides by 2 to obtain π radians = 180◦ . Now divide both sides by 9, obtaining 180 ◦ π radians = = 20◦ . 9 9 Exercise 11 Student Solution Manual, Section 5.2 Exercise 13 13. 3 radians solution Start with the equation 2π radians = 360◦ . Divide both sides by 2π to obtain 1 radian = 180 ◦ . π Now multiply both sides by 3, obtaining 3 radians = 3 · 540 ◦ 180 ◦ = . π π Instructor’s Solutions Manual, Section 5.2 15. − 2π 3 radians solution Start with the equation 2π radians = 360◦ . Divide both sides by 2 to obtain π radians = 180◦ . 2 Now multiply both sides by − 3 , obtaining − 2 2π radians = − · 180◦ = −120◦ . 3 3 Exercise 15 Student Solution Manual, Section 5.2 Exercise 17 17. Suppose an ant walks counterclockwise on the unit circle from the point 5π (0, 1) to the endpoint of the radius that forms an angle of 4 radians with the positive horizontal axis. How far has the ant walked? solution The radius whose endpoint equals (0, 1) makes an angle of π 2 radians with the positive horizontal axis. This radius corresponds to the smaller angle shown below. 5π Because 4 = π + π4 , the radius that forms an angle of 5π 4 radians with π the positive horizontal axis lies 4 radians beyond the negative horizontal axis (half-way between the negative horizontal axis and the negative vertical axis). Thus the ant ends its walk at the endpoint of the radius corresponding to the larger angle shown below: 1 The ant walks along the thickened circular arc shown above. This circular 5π π 3π arc corresponds to an angle of 4 − 2 radians, which equals 4 radians. 3π Thus the distance walked by the ant is 4 . Instructor’s Solutions Manual, Section 5.2 19. Exercise 19 Find the lengths of both circular arcs of the unit circle connecting the point (1, 0) and the endpoint of the radius that makes an angle of 3 radians with the positive horizontal axis. solution Because 3 is a bit less than π , the radius that makes an angle of 3 radians with the positive horizontal axis lies a bit above the negative horizontal axis, as shown below. The thickened circular arc corresponds to an angle of 3 radians and thus has length 3. The entire unit circle has length 2π . Thus the length of the other circular arc is 2π − 3, which is approximately 3.28. 1 Student Solution Manual, Section 5.2 21. Exercise 21 Find √the lengths of both circular arcs of the unit circle connecting the √ 2 2 point 2 , − 2 and the point whose radius makes an angle of 1 radian with the positive horizontal axis. √ √2 2 solution The radius of the unit circle whose endpoint equals 2 , − 2 π makes an angle of − 4 radians with the positive horizontal axis, as shown with the clockwise arrow below. The radius that makes an angle of 1 radian with the positive horizontal axis is shown with a counterclockwise arrow. 1 π Thus the thickened circular arc above corresponds to an angle of 1 + 4 and thus has length 1 + π4 , which is approximately 1.79. The entire unit circle has length 2π . Thus the length of the other circular arc below is π 7π 2π − (1 + 4 ), which equals 4 − 1, which is approximately 4.50. Instructor’s Solutions Manual, Section 5.2 23. For a 16-inch pizza, ﬁnd the area of a slice with angle Exercise 23 3 4 radians. solution Pizzas are measured by their diameters; thus this pizza has 1 3 a radius of 8 inches. Thus the area of the slice is 2 · 4 · 82 , which equals 24 square inches. Student Solution Manual, Section 5.2 Exercise 25 25. Suppose a slice of a 12-inch pizza has an area of 20 square inches. What is the angle of this slice? solution This pizza has a radius of 6 inches. Let θ denote the angle of this slice, measured in radians. Then 1 20 = 2 θ · 62 . Solving this equation for θ, we get θ = 10 9 radians. Instructor’s Solutions Manual, Section 5.2 27. Exercise 27 Suppose a slice of pizza with an angle of 56 radians has an area of 21 square inches. What is the diameter of this pizza? solution Let r denote the radius of this pizza. Thus 21 = 1 2 · 56 r 2 . 252 Solving this equation for r , we get r = 5 ≈ 7.1. Thus the diameter of the pizza is approximately 14.2 inches. Student Solution Manual, Section 5.2 Exercise 29 For each of the angles in Exercises 29–34, ﬁnd the endpoint of the radius of the unit circle that makes the given angle with the positive horizontal axis. 29. 5π 6 radians solution For this exercise it may be easier to convert to degrees. Thus 5π we translate 6 radians to 150◦ . The radius making a 150◦ angle with the positive horizontal axis is shown below. The angle from this radius to the negative horizontal axis equals 180◦ − 150◦ , which equals 30◦ as shown in the ﬁgure below. Drop a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 30◦ ; thus the other angle must be 60◦ , as labeled below. The side of the right triangle opposite the 30◦ angle has length ◦ √ 3 2 . 1 2; the Looking side of the right triangle opposite the 60 angle has length at the ﬁgure below, we see that the ﬁrst coordinate of the endpoint of the radius is the negative of the length of the side opposite the 60◦ angle, and the second coordinate of the endpoint of the radius is the length of the side opposite the 30◦ angle. Thus the endpoint of the radius is √3 1 − 2 ,2 . Instructor’s Solutions Manual, Section 5.2 Exercise 29 150 60 30 1 Student Solution Manual, Section 5.2 Exercise 31 31. − π4 radians solution For this exercise it may be easier to convert to degrees. Thus π we translate − 4 radians to −45◦ . The radius making a −45◦ angle with the positive horizontal axis is shown below. Draw a perpendicular line segment from the endpoint of the radius to the horizontal axis, forming a right triangle as shown below. We already know that one angle of this right triangle is 45◦ ; thus the other angle must also be 45◦ , as labeled below. The hypotenuse of this right triangle is a radius of the unit circle and √ 2 thus has length 1. The other two sides each have length 2 . Looking at the ﬁgure√below, we see that the ﬁrst coordinate of the endpoint of the 2 radius is 2 and the second coordinate of the endpoint of the radius is √ √ √2 2 2 − 2 . Thus the endpoint of the radius is 2 , − 2 . 45 1 45 Instructor’s Solutions Manual, Section 5.2 33. 5π 2 Exercise 33 radians 5π π solution Note that 2 = 2π + 2 . Thus the radius making an angle of 5π 2 radians with the positive horizontal axis is obtained by starting at the horizontal axis, making one complete counterclockwise rotation (which π is 2π radians), and then continuing for another 2 radians. The resulting radius is shown below. Its endpoint is (0, 1). 1 Student Solution Manual, Section 5.3 Exercise 1 Solutions to Exercises, Section 5.3 Give exact values for the quantities in Exercises 1–10. Do not use a calculator for any of these exercises—otherwise you will likely get decimal approximations for some solutions rather than exact answers. More importantly, good understanding will come from working these exercises by hand. 1. (a) cos 3π (b) sin 3π solution Because 3π = 2π + π , an angle of 3π radians (as measured counterclockwise from the positive horizontal axis) consists of a complete revolution around the circle (2π radians) followed by another π radians (180◦ ), as shown below. The endpoint of the corresponding radius is (−1, 0). Thus cos 3π = −1 and sin 3π = 0. 1 Instructor’s Solutions Manual, Section 5.3 3. (a) cos 11π 4 Exercise 3 (b) sin 11π 4 solution Because 11π = 2π + π2 + π4 , an angle of 11π radians (as 4 4 measured counterclockwise from the positive horizontal axis) consists of a complete revolution around the circle (2π radians) followed by anπ π other 2 radians (90◦ ), followed by another 4 radians (45◦ ), as shown √2 √2 below. Hence the endpoint of the corresponding radius is − 2 , 2 . Thus cos 11π 4 =− √ 2 2 and sin 11π 4 = √ 2 2 . 1 Student Solution Manual, Section 5.3 5. (a) cos 2π 3 Exercise 5 (b) sin 2π 3 π π 2π solution Because 2π 3 = 2 + 6 , an angle of 3 radians (as measured π counterclockwise from the positive horizontal axis) consists of 2 radians π ◦ ◦ (90 radians) followed by another 6 radians (30 ), as shown below. The 1 √3 2π 1 endpoint of the corresponding radius is − 2 , 2 . Thus cos 3 = − 2 and sin 2π 3 = √ 3 2 . 1 Instructor’s Solutions Manual, Section 5.3 7. (a) cos 210◦ Exercise 7 (b) sin 210◦ solution Because 210 = 180 + 30, an angle of 210◦ (as measured counterclockwise from the positive horizontal axis) consists of 180◦ followed ◦ , as shown below. The √endpoint of the corresponding by another 30 √3 1 3 1 radius is − 2 , − 2 . Thus cos 210◦ = − 2 and sin 210◦ = − 2 . 1 Student Solution Manual, Section 5.3 9. Exercise 9 (a) cos 360045◦ (b) sin 360045◦ solution Because 360045 = 360 × 1000 + 45, an angle of 360045◦ (as measured counterclockwise from the positive horizontal axis) consists of 1000 complete revolutions around the circle√followed by another 45◦ . 2 √2 The endpoint of the corresponding radius is 2 , 2 . Thus cos 360045◦ = √ 2 2 and sin 360045◦ = √ 2 2 . Instructor’s Solutions Manual, Section 5.3 Exercise 11 11. Find the smallest number θ larger than 4π such that cos θ = 0. solution Note that 0 = cos π 2 = cos 3π 2 = cos 5π 2 = ... and that the only numbers whose cosine equals 0 are of the form (2n+1)π , 2 where n is an integer. The smallest number of this form larger than 4π is 9π 9π 2 . Thus 2 is the smallest number larger than 4π whose cosine equals 0. Student Solution Manual, Section 5.3 Exercise 13 13. Find the four smallest positive numbers θ such that cos θ = 0. solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, the ﬁrst coordinate of its endpoint ﬁrst becomes 0 π 3π when θ equals 2 (which equals 90◦ ), then again when θ equals 2 (which 5π equals 270◦ ), then again when θ equals 2 (which equals 360◦ + 90◦ , or 7π 450◦ ), then again when θ equals 2 (which equals 360◦ + 270◦ , or 630◦ ), and so on. Thus the four smallest positive numbers θ such that cos θ = 0 π 3π 5π 7π are 2 , 2 , 2 , and 2 . Instructor’s Solutions Manual, Section 5.3 Exercise 15 15. Find the four smallest positive numbers θ such that sin θ = 1. solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, then the second coordinate of its endpoint ﬁrst π becomes 1 when θ equals 2 (which equals 90◦ ), then again when θ equals 5π 9π ◦ ◦ ◦ 2 (which equals 360 +90 , or 450 ), then again when θ equals 2 (which 13π equals 2 × 360◦ + 90◦ , or 810◦ ), then again when θ equals 2 (which ◦ ◦ ◦ equals 3 × 360 + 90 , or 1170 ), and so on. Thus the four smallest π 5π 9π 13π positive numbers θ such that sin θ = 1 are 2 , 2 , 2 , and 2 . Student Solution Manual, Section 5.3 Exercise 17 17. Find the four smallest positive numbers θ such that cos θ = −1. solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, the ﬁrst coordinate of its endpoint ﬁrst becomes −1 when θ equals π (which equals 180◦ ), then again when θ equals 3π (which equals 360◦ + 180◦ , or 540◦ ), then again when θ equals 5π (which equals 2 × 360◦ + 180◦ , or 900◦ ), then again when θ equals 7π (which equals 3 × 360◦ + 180◦ , or 1260◦ ), and so on. Thus the four smallest positive numbers θ such that cos θ = −1 are π , 3π , 5π , and 7π . Instructor’s Solutions Manual, Section 5.3 Exercise 19 19. Find the four smallest positive numbers θ such that sin θ = 12 . solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the posi1 tive horizontal axis, the second coordinate of its endpoint ﬁrst becomes 2 π 5π when θ equals 6 (which equals 30◦ ), then again when θ equals 6 (which 13π equals 150◦ ), then again when θ equals 6 (which equals 360◦ + 30◦ , or 17π 390◦ ), then again when θ equals 6 (which equals 360◦ + 150◦ , or 510◦ ), 1 and so on. Thus the four smallest positive numbers θ such that sin θ = 2 π 5π 13π 17π are 6 , 6 , 6 , and 6 . Student Solution Manual, Section 5.3 21. Suppose 0 < θ < π 2 Exercise 21 and cos θ = 25 . Evaluate sin θ. solution We know that (cos θ)2 + (sin θ)2 = 1. Thus (sin θ)2 = 1 − (cos θ)2 =1− = 2 2 5 21 . 25 Because 0 < θ < π2 , we know that sin θ > 0. Thus taking square roots of both sides of the equation above gives √ 21 . sin θ = 5 Instructor’s Solutions Manual, Section 5.3 23. Suppose π 2 Exercise 23 < θ < π and sin θ = 29 . Evaluate cos θ. solution We know that (cos θ)2 + (sin θ)2 = 1. Thus (cos θ)2 = 1 − (sin θ)2 =1− = 2 2 9 77 . 81 Because π2 < θ < π , we know that cos θ < 0. Thus taking square roots of both sides of the equation above gives √ 77 . cos θ = − 9 Student Solution Manual, Section 5.3 25. Exercise 25 Suppose − π2 < θ < 0 and cos θ = 0.1. Evaluate sin θ. solution We know that (cos θ)2 + (sin θ)2 = 1. Thus (sin θ)2 = 1 − (cos θ)2 = 1 − (0.1)2 = 0.99. π Because − 2 < θ < 0, we know that sin θ < 0. Thus taking square roots of both sides of the equation above gives √ sin θ = − 0.99 ≈ −0.995. Instructor’s Solutions Manual, Section 5.3 Exercise 27 27. Find the smallest number x such that sin(ex ) = 0. solution Note that ex is an increasing function. Because ex is positive for every real number x, and because π is the smallest positive number whose sine equals 0, we want to choose x so that ex = π . Thus x = ln π . Student Solution Manual, Section 5.3 29. Exercise 29 Find the smallest positive number x such that sin(x 2 + x + 4) = 0. solution Note that x 2 + x + 4 is an increasing function on the interval [0, ∞). If x is positive, then x 2 + x + 4 > 4. Because 4 is larger than π but less than 2π , the smallest number bigger than 4 whose sine equals 0 is 2π . Thus we want to choose x so that x 2 + x + 4 = 2π . In other words, we need to solve the equation x 2 + x + (4 − 2π ) = 0. Using the quadratic formula, we see that the solutions to this equation are √ −1 ± 8π − 15 . x= 2 A calculator shows that choosing the plus sign in the equation above gives x ≈ 1.0916 and choosing the minus sign gives x ≈ −2.0916. We seek only positive values of x, and thus we choose the plus sign in the equation above, getting x ≈ 1.0916. Instructor’s Solutions Manual, Section 5.4 Exercise 1 Solutions to Exercises, Section 5.4 1. Find the four smallest positive numbers θ such that tan θ = 1. solution Think of a radius of the unit circle whose endpoint is (1, 0). If this radius moves counterclockwise, forming an angle of θ with the positive horizontal axis, then the ﬁrst and second coordinates of its endpoint ﬁrst become equal (which is equivalent to having tan θ = 1) when θ π 5π equals 4 (which equals 45◦ ), then again when θ equals 4 (which equals 9π ◦ ◦ 225 ), then again when θ equals 4 (which equals 360 + 45◦ , or 405◦ ), 13π then again when θ equals 4 (which equals 360◦ + 225◦ , or 585◦ ), and so on. Thus the four smallest positive numbers θ such that tan θ = 1 are 9π 13π 4 , and 4 . π 5π 4, 4 , Student Solution Manual, Section 5.4 3. Suppose 0 < θ < π 2 Exercise 3 and cos θ = 15 . Evaluate: (a) sin θ (b) tan θ solution The ﬁgure below gives a sketch of the angle involved in this exercise: Θ 1 The angle between 0 and 1 equals 5 . π 2 whose cosine (a) We know that (cos θ)2 + (sin θ)2 = 1. Thus 1 2 5 + (sin θ)2 = 1. Solving this equation for (sin θ)2 gives (sin θ)2 = 24 . 25 The sketch above shows that sin θ > 0. Thus taking square roots of both sides of the equation above gives Instructor’s Solutions Manual, Section 5.4 Exercise 3 √ √ √ 2 6 24 4·6 = = . sin θ = 5 5 5 (b) sin θ = tan θ = cos θ √ 2 6 5 1 5 √ = 2 6. Student Solution Manual, Section 5.4 5. Suppose π 2 Exercise 5 < θ < π and sin θ = 23 . Evaluate: (a) cos θ (b) tan θ solution The ﬁgure below gives a sketch of the angle involved in this exercise: Θ 1 The angle between 2 equals 3 . π 2 and π whose sine (a) We know that (cos θ)2 + (sin θ)2 = 1. Thus (cos θ)2 + 2 2 3 = 1. Solving this equation for (cos θ)2 gives (cos θ)2 = 5 . 9 The sketch above shows that cos θ < 0. Thus taking square roots of both sides of the equation above gives Instructor’s Solutions Manual, Section 5.4 Exercise 5 √ 5 . cos θ = − 3 (b) √ 2 2 5 2 sin θ 3 = − √5 = − √ = − . tan θ = cos θ 5 5 3 Student Solution Manual, Section 5.4 Exercise 7 7. Suppose − π2 < θ < 0 and cos θ = 45 . Evaluate: (a) sin θ (b) tan θ solution The ﬁgure below gives a sketch of the angle involved in this exercise: Θ 1 The angle between − π2 and 0 whose 4 cosine equals 5 . (a) We know that (cos θ)2 + (sin θ)2 = 1. Thus 4 2 5 + (sin θ)2 = 1. Solving this equation for (sin θ)2 gives (sin θ)2 = 9 . 25 The sketch above shows that sin θ < 0. Thus taking square roots of both sides of the equation above gives Instructor’s Solutions Manual, Section 5.4 Exercise 7 3 sin θ = − . 5 (b) 3 tan θ = 3 sin θ = − 54 = − . cos θ 4 5 Student Solution Manual, Section 5.4 9. Suppose 0 < θ < π 2 Exercise 9 and tan θ = 14 . Evaluate: (a) cos θ (b) sin θ solution The ﬁgure below gives a sketch of the angle involved in this exercise: Θ 1 The angle between 0 and 1 tangent equals 4 . (a) Rewrite the equation tan θ = 14 in the form sides of this equation by cos θ, we get sin θ = 1 4 sin θ cos θ = 1 4. π 2 whose Multiplying both cos θ. Substitute this expression for sin θ into the equation (cos θ)2 +(sin θ)2 = 1, getting 1 (cos θ)2 + 16 (cos θ)2 = 1, which is equivalent to Instructor’s Solutions Manual, Section 5.4 Exercise 9 (cos θ)2 = 16 . 17 The sketch above shows that cos θ > 0. Thus taking square roots of both sides of the equation above gives √ 4 17 4 = . cos θ = √ 17 17 (b) We have already noted that sin θ = 1 4 cos θ. Thus √ 17 . sin θ = 17 Student Solution Manual, Section 5.4 Exercise 11 11. Suppose − π2 < θ < 0 and tan θ = −3. Evaluate: (a) cos θ (b) sin θ solution The ﬁgure below gives a sketch of the angle involved in this exercise: Θ 1 The angle between − π2 and 0 whose tangent equals −3. (a) Rewrite the equation tan θ = −3 in the form sides of this equation by cos θ, we get sin θ cos θ = −3. Multiplying both sin θ = −3 cos θ. Substitute this expression for sin θ into the equation (cos θ)2 +(sin θ)2 = 1, getting (cos θ)2 + 9(cos θ)2 = 1, which is equivalent to Instructor’s Solutions Manual, Section 5.4 (cos θ)2 = Exercise 11 1 . 10 The sketch above shows that cos θ > 0. Thus taking square roots of both sides of the equation above gives √ 10 1 = . cos θ = √ 10 10 (b) We have already noted that sin θ = −3 cos θ. Thus √ 3 10 . sin θ = − 10 Student Solution Manual, Section 5.4 Given that cos 15◦ = √ 2+ 3 2 Exercise 13 and sin 22.5◦ = √ 2− 2 2 , in Exercises 13–22 ﬁnd exact expressions for the indicated quantities. [These values for cos 15◦ and sin 22.5◦ will be derived in Examples 4 and 5 in Section 6.3.] 13. sin 15◦ solution We know that (cos 15◦ )2 + (sin 15◦ )2 = 1. Thus (sin 15◦ )2 = 1 − (cos 15◦ )2 2 + √3 2 =1− 2 √ 2+ 3 =1− 4 √ 2− 3 . = 4 Because sin 15◦ > 0, taking square roots of both sides of the equation above gives √ 2− 3 ◦ . sin 15 = 2 Instructor’s Solutions Manual, Section 5.4 15. tan 15◦ solution sin 15◦ cos 15◦ √ 2− 3 = √ 2+ 3 √ √ 2− 3 2− 3 · = √ √ 2+ 3 2− 3 √ 2− 3 √ = 4−3 √ =2− 3 tan 15◦ = Exercise 15 Student Solution Manual, Section 5.4 Exercise 17 17. cot 15◦ solution cot 15◦ = 1 tan 15◦ = 1 √ 2− 3 √ 2+ 3 1 √ · √ = 2− 3 2+ 3 √ 2+ 3 = 4−3 √ =2+ 3 Instructor’s Solutions Manual, Section 5.4 Exercise 19 19. csc 15◦ solution csc 15◦ = 1 sin 15◦ = 2 2− √ 3 √ 2+ 3 √ · √ 2− 3 2+ 3 √ 2 2+ 3 = √ 4−3 √ =2 2+ 3 = 2 Student Solution Manual, Section 5.4 Exercise 21 21. sec 15◦ solution sec 15◦ = 1 cos 15◦ 2 = √ 2+ 3 √ 2− 3 2 = √ · √ 2+ 3 2− 3 √ 2 2− 3 = √ 4−3 √ =2 2− 3 Instructor’s Solutions Manual, Section 5.4 Exercise 23 π 2 ), Suppose u and ν are in the interval (0, tan u = 2 and with tan ν = 3. In Exercises 23–32, ﬁnd exact expressions for the indicated quantities. 23. cot u solution cot u = = 1 tan u 1 2 Student Solution Manual, Section 5.4 Exercise 25 25. cos u solution We know that 2 = tan u = sin u . cos u To ﬁnd cos u, make the substitution sin u = 1 − (cos u)2 in the equaπ tion above (this substitution is valid because we know that 0 < u < 2 and thus sin u > 0), getting 1 − (cos u)2 . 2= cos u Now square both sides of the equation above, then multiply both sides by (cos u)2 and rearrange to get the equation 5(cos u)2 = 1. π Because 0 < u < 2 , we see that cos u > 0. Thus taking square roots of 1 both sides of the equation above gives cos u = √5 , which can be rewritten as cos u = √ 5 5 . Instructor’s Solutions Manual, Section 5.4 27. sin u solution sin u = 1 − (cos u)2 1 = 1− 5 4 = 5 2 = √ 5 √ 2 5 = 5 Exercise 27 Student Solution Manual, Section 5.4 Exercise 29 29. csc u solution 1 sin u √ 5 = 2 csc u = Instructor’s Solutions Manual, Section 5.4 Exercise 31 31. sec u solution 1 cos u √ = 5 sec u = Student Solution Manual, Section 5.4 33. Exercise 33 Find the smallest number x such that tan ex = 0. solution Note that ex is an increasing function. Because ex is positive for every real number x, and because π is the smallest positive number whose tangent equals 0, we want to choose x so that ex = π . Thus x = ln π ≈ 1.14473. Instructor’s Solutions Manual, Section 5.5 Exercise 1 Solutions to Exercises, Section 5.5 Use the right triangle below for Exercises 1–76. This triangle is not drawn to scale corresponding to the data in the exercises. c Ν b u a 1. Suppose a = 2 and b = 7. Evaluate c. solution The Pythagorean Theorem implies that c 2 = 22 + 72 . Thus √ c = 22 + 72 = 53. Student Solution Manual, Section 5.5 3. Suppose a = 2 and b = 7. Evaluate cos u. solution √ 2 53 a 2 adjacent side √ = = = cos u = 53 hypotenuse c 53 Exercise 3 Instructor’s Solutions Manual, Section 5.5 5. Suppose a = 2 and b = 7. Evaluate sin u. solution √ 7 53 b 7 opposite side √ = = = sin u = 53 hypotenuse c 53 Exercise 5 Student Solution Manual, Section 5.5 7. Suppose a = 2 and b = 7. Evaluate tan u. solution tan u = b 7 opposite side = = adjacent side a 2 Exercise 7 Instructor’s Solutions Manual, Section 5.5 9. Suppose a = 2 and b = 7. Evaluate cos ν. solution √ 7 53 b 7 adjacent side √ = = = cos ν = 53 hypotenuse c 53 Exercise 9 Student Solution Manual, Section 5.5 11. Suppose a = 2 and b = 7. Evaluate sin ν. solution √ 2 53 a 2 opposite side √ = = = sin ν = 53 hypotenuse c 53 Exercise 11 Instructor’s Solutions Manual, Section 5.5 13. Suppose a = 2 and b = 7. Evaluate tan ν. solution tan ν = a 2 opposite side = = adjacent side b 7 Exercise 13 Student Solution Manual, Section 5.5 Exercise 15 15. Suppose b = 2 and c = 7. Evaluate a. solution The Pythagorean Theorem implies that a2 + 22 = 72 . Thus √ √ √ √ √ a = 72 − 22 = 45 = 9 · 5 = 9 · 5 = 3 5. Instructor’s Solutions Manual, Section 5.5 17. Suppose b = 2 and c = 7. Evaluate cos u. solution √ a 3 5 adjacent side = = cos u = hypotenuse c 7 Exercise 17 Student Solution Manual, Section 5.5 19. Suppose b = 2 and c = 7. Evaluate sin u. solution sin u = b 2 opposite side = = hypotenuse c 7 Exercise 19 Instructor’s Solutions Manual, Section 5.5 21. Suppose b = 2 and c = 7. Evaluate tan u. solution √ 2 5 b 2 opposite side √ = = = tan u = 15 adjacent side a 3 5 Exercise 21 Student Solution Manual, Section 5.5 23. Suppose b = 2 and c = 7. Evaluate cos ν. solution cos ν = b 2 adjacent side = = hypotenuse c 7 Exercise 23 Instructor’s Solutions Manual, Section 5.5 25. Suppose b = 2 and c = 7. Evaluate sin ν. solution √ a 3 5 opposite side = = sin ν = hypotenuse c 7 Exercise 25 Student Solution Manual, Section 5.5 27. Suppose b = 2 and c = 7. Evaluate tan ν. solution √ a 3 5 opposite side = = tan ν = adjacent side b 2 Exercise 27 Instructor’s Solutions Manual, Section 5.5 29. Suppose a = 5 and u = 17◦ . Evaluate b. solution We have tan 17◦ = opposite side b = . adjacent side 5 Solving for b, we get b = 5 tan 17◦ ≈ 1.53. Exercise 29 Student Solution Manual, Section 5.5 31. Exercise 31 Suppose a = 5 and u = 17◦ . Evaluate c. solution We have cos 17◦ = 5 adjacent side = . hypotenuse c Solving for c, we get c= 5 ≈ 5.23. cos 17◦ Instructor’s Solutions Manual, Section 5.5 33. Exercise 33 Suppose u = 17◦ . Evaluate cos ν. solution Because ν = 90◦ −u, we have ν = 73◦ . Thus cos ν = cos 73◦ ≈ 0.292. Student Solution Manual, Section 5.5 35. Suppose u = 17◦ . Evaluate sin ν. solution sin ν = sin 73◦ ≈ 0.956 Exercise 35 Instructor’s Solutions Manual, Section 5.5 37. Suppose u = 17◦ . Evaluate tan ν. solution tan ν = tan 73◦ ≈ 3.27 Exercise 37 Student Solution Manual, Section 5.5 39. Exercise 39 Suppose c = 8 and u = 1 radian. Evaluate a. solution We have cos 1 = a adjacent side = . hypotenuse 8 Solving for a, we get a = 8 cos 1 ≈ 4.32. When using a calculator to do the approximation above, be sure that your calculator is set to operate in radian mode. Instructor’s Solutions Manual, Section 5.5 41. Suppose c = 8 and u = 1 radian. Evaluate b. solution We have sin 1 = b opposite side = . hypotenuse 8 Solving for b, we get b = 8 sin 1 ≈ 6.73. Exercise 41 Student Solution Manual, Section 5.5 43. Exercise 43 Suppose u = 1 radian. Evaluate cos ν. solution Because ν = π 2 − u, we have ν = π π 2 − 1. Thus cos ν = cos( 2 − 1) ≈ 0.841. Instructor’s Solutions Manual, Section 5.5 45. Exercise 45 Suppose u = 1 radian. Evaluate sin ν. solution π sin ν = sin( 2 − 1) ≈ 0.540 Student Solution Manual, Section 5.5 47. Exercise 47 Suppose u = 1 radian. Evaluate tan ν. solution π tan ν = tan( 2 − 1) ≈ 0.642 Instructor’s Solutions Manual, Section 5.5 Exercise 49 49. Suppose c = 4 and cos u = 15 . Evaluate a. solution We have adjacent side a 1 = cos u = = . 5 hypotenuse 4 Solving this equation for a, we get a= 4 . 5 Student Solution Manual, Section 5.5 Exercise 51 51. Suppose c = 4 and cos u = 15 . Evaluate b. solution The Pythagorean Theorem implies that b= 16 =4 16 − 25 1 =4 1− 25 4 2 5 + b2 = 42 . Thus √ 8 6 24 = . 25 5 Instructor’s Solutions Manual, Section 5.5 Exercise 53 53. Suppose cos u = 15 . Evaluate sin u. solution sin u = 1 − (cos u)2 = 1− = 1 25 √ 24 2 6 = 25 5 Student Solution Manual, Section 5.5 Exercise 55 55. Suppose cos u = 15 . Evaluate tan u. solution sin u = tan u = cos u √ 2 6 5 1 5 √ =2 6 Instructor’s Solutions Manual, Section 5.5 57. Suppose cos u = 15 . Evaluate cos ν. solution √ 2 6 b cos ν = = sin u = c 5 Exercise 57 Student Solution Manual, Section 5.5 Exercise 59 59. Suppose cos u = 15 . Evaluate sin ν. solution sin ν = 1 a = cos u = c 5 Instructor’s Solutions Manual, Section 5.5 Exercise 61 61. Suppose cos u = 15 . Evaluate tan ν. solution sin ν = tan ν = cos ν 1 5 √ 2 6 5 √ 6 1 = √ = 12 2 6 Student Solution Manual, Section 5.5 Exercise 63 63. Suppose b = 4 and sin ν = 13 . Evaluate a. solution We have opposite side a 1 = sin ν = = . 3 hypotenuse c Thus c = 3a. By the Pythagorean Theorem, we also have c 2 = a2 + 16. Substituting 3a for c in this equation gives 9a2 = a2 + 16. Solving the equation above for a shows that a = √ 2. Instructor’s Solutions Manual, Section 5.5 65. Suppose b = 4 and sin ν = 13 . Evaluate c. solution We have Thus a 1 = sin ν = . 3 c √ c = 3a = 3 2. Exercise 65 Student Solution Manual, Section 5.5 Exercise 67 67. Suppose sin ν = 13 . Evaluate cos u. solution cos u = 1 a = sin ν = c 3 Instructor’s Solutions Manual, Section 5.5 Exercise 69 69. Suppose sin ν = 13 . Evaluate sin u. solution sin u = 1 − (cos u)2 = 1− = 1 2 3 √ 8 2 2 = 9 3 Student Solution Manual, Section 5.5 Exercise 71 71. Suppose sin ν = 13 . Evaluate tan u. solution sin u = tan u = cos u √ 2 2 3 1 3 √ =2 2 Instructor’s Solutions Manual, Section 5.5 Exercise 73 73. Suppose sin ν = 13 . Evaluate cos ν. solution cos ν = 1 − (sin ν)2 = 1− = 1 2 3 √ 8 2 2 = 9 3 Student Solution Manual, Section 5.5 Exercise 75 75. Suppose sin ν = 13 . Evaluate tan ν. solution sin ν = tan ν = cos ν 1 3 √ 2 2 3 √ 2 1 = √ = 4 2 2 Instructor’s Solutions Manual, Section 5.5 77. Exercise 77 Suppose a 25-foot ladder is leaning against a wall, making a 63◦ angle with the ground (as measured from a perpendicular line from the base of the ladder to the wall). How high up the wall is the end of the ladder? solution In the sketch here, the vertical line represents the wall and the hypotenuse represents the ladder. As labeled here, the ladder touches the wall at height b; thus we need to evaluate b. 25 b 63 We have sin 63◦ = b 25 . Solving this equation for b, we get b = 25 sin 63◦ ≈ 22.28. Thus the ladder touches the wall at a height of approximately 22.28 feet. Because 0.28 × 12 = 3.36, this is approximately 22 feet, 3 inches. Student Solution Manual, Section 5.5 79. Exercise 79 Suppose you need to ﬁnd the height of a tall building. Standing 20 meters away from the base of the building, you aim a laser pointer at the closest part of the top of the building. You measure that the laser pointer is 4◦ tilted from pointing straight up. The laser pointer is held 2 meters above the ground. How tall is the building? solution In the sketch here, the rightmost vertical line represents the building and the hypotenuse represents the path of the laser beam. Because the laser pointer is 4◦ tilted from pointing straight up, the b 2 86 20 angle formed by the laser beam and a line parallel to the ground is 86◦ , as indicated in the ﬁgure (which is not drawn to scale). The side of the right triangle opposite the 86◦ angle has been labeled b. Thus the height of the building is b + 2. We have tan 86◦ = b 20 . Solving this equation for b, we get b = 20 tan 86◦ ≈ 286. Adding 2 to this result, we see that the height of the building is approximately 288 meters. Instructor’s Solutions Manual, Section 5.6 Exercise 1 Solutions to Exercises, Section 5.6 1. For θ = 7◦ , evaluate each of the following: (a) cos2 θ (b) cos(θ 2 ) [Exercises 1 and 2 emphasize that cos2 θ does not equal cos(θ 2 ).] solution (a) Using a calculator working in degrees, we have cos2 7◦ = (cos 7◦ )2 ≈ (0.992546)2 ≈ 0.985148. (b) Note that 72 = 49. Using a calculator working in degrees, we have cos 49◦ ≈ 0.656059. Student Solution Manual, Section 5.6 3. Exercise 3 For θ = 4 radians, evaluate each of the following: (a) sin2 θ (b) sin(θ 2 ) [Exercises 3 and 4 emphasize that sin2 θ does not equal sin(θ 2 ).] solution (a) Using a calculator working in radians, we have sin2 4 = (sin 4)2 ≈ (−0.756802)2 ≈ 0.57275. (b) Note that 42 = 16. Using a calculator working in radians, we have sin 16 ≈ −0.287903. Instructor’s Solutions Manual, Section 5.6 Exercise 5 In Exercises 5–38, ﬁnd exact expressions for the indicated quantities, given that √ √ 2 + 3 2− 2 π π and sin 8 = . cos 12 = 2 2 [These values for cos Section 6.3.] π 12 and sin π 8 will be derived in Examples 4 and 5 in π 5. cos(− 12 ) solution π cos(− 12 ) = cos π 12 √ 2+ 3 = 2 Student Solution Manual, Section 5.6 Exercise 7 π 7. sin 12 solution We know that π π cos2 12 + sin2 12 = 1. Thus π π sin2 12 = 1 − cos2 12 2 + √3 2 =1− 2 √ 2+ 3 =1− 4 √ 2− 3 . = 4 π Because sin 12 > 0, taking square roots of both sides of the equation above gives √ 2− 3 π . sin 12 = 2 Instructor’s Solutions Manual, Section 5.6 Exercise 9 π 9. sin(− 12 ) solution π sin(− 12 ) = − sin π 12 √ 2− 3 =− 2 Student Solution Manual, Section 5.6 Exercise 11 π 11. tan 12 solution π tan 12 = π sin 12 π cos 12 √ 2− 3 = √ 2+ 3 √ √ 2− 3 2− 3 = √ · √ 2+ 3 2− 3 √ 2− 3 = √ 4−3 √ =2− 3 Instructor’s Solutions Manual, Section 5.6 Exercise 13 π 13. tan(− 12 ) solution π π tan(− 12 ) = − tan 12 = −(2 − √ 3) = √ 3−2 Student Solution Manual, Section 5.6 Exercise 15 15. cos 25π 12 solution Because 25π 12 = π 12 cos + 2π , we have 25π 12 π = cos( 12 + 2π ) π = cos 12 √ 2+ 3 . = 2 Instructor’s Solutions Manual, Section 5.6 Exercise 17 17. sin 25π 12 solution Because 25π 12 = π 12 sin + 2π , we have 25π 12 π = sin( 12 + 2π ) π = sin 12 √ 2− 3 . = 2 Student Solution Manual, Section 5.6 Exercise 19 19. tan 25π 12 solution Because 25π 12 = π 12 tan + 2π , we have 25π 12 π = tan( 12 + 2π ) π = tan 12 √ = 2 − 3. Instructor’s Solutions Manual, Section 5.6 Exercise 21 21. cos 13π 12 solution Because 13π 12 = π 12 cos + π , we have 13π 12 π = cos( 12 + π ) π = − cos 12 √ 2+ 3 . =− 2 Student Solution Manual, Section 5.6 Exercise 23 23. sin 13π 12 solution Because 13π 12 = π 12 sin + π , we have 13π 12 π = sin( 12 + π ) π = − sin 12 √ 2− 3 . =− 2 Instructor’s Solutions Manual, Section 5.6 Exercise 25 25. tan 13π 12 solution Because 13π 12 = π 12 tan + π , we have 13π 12 π = tan( 12 + π ) π = tan 12 √ = 2 − 3. Student Solution Manual, Section 5.6 Exercise 27 27. cos 5π 12 solution cos 5π 12 = π sin( 2 − 5π 12 ) = sin π 12 √ 2− 3 = 2 Instructor’s Solutions Manual, Section 5.6 Exercise 29 29. cos(− 5π 12 ) solution 5π cos(− 12 ) = cos 5π 12 √ 2− 3 = 2 Student Solution Manual, Section 5.6 Exercise 31 31. sin 5π 12 solution sin 5π 12 = π cos( 2 − 5π 12 ) = cos π 12 √ 2+ 3 = 2 Instructor’s Solutions Manual, Section 5.6 Exercise 33 33. sin(− 5π 12 ) solution 5π sin(− 12 ) = − sin 5π 12 √ 2+ 3 =− 2 Student Solution Manual, Section 5.6 Exercise 35 35. tan 5π 12 solution tan 5π 12 = 1 tan( π2 − = 1 π tan 12 = 1 √ 2− 3 5π 12 ) √ 2+ 3 1 √ · √ = 2− 3 2+ 3 √ 2+ 3 = 4−3 √ =2+ 3 Instructor’s Solutions Manual, Section 5.6 Exercise 37 37. tan(− 5π 12 ) solution 5π tan(− 12 ) = − tan 5π 12 = −2 − √ 3 Student Solution Manual, Section 5.6 Exercise 39 Suppose u and ν are in the interval ( π 2 , π), with tan u = −2 and tan ν = −3. In Exercises 39–66, ﬁnd exact expressions for the indicated quantities. 39. tan(−u) solution tan(−u) = − tan u = −(−2) = 2 Instructor’s Solutions Manual, Section 5.6 Exercise 41 41. cos u solution We know that −2 = tan u = sin u . cos u √ To ﬁnd cos u, make the substitution sin u = 1 − cos2 u in the equation π above (this substitution is valid because 2 < u < π , which implies that sin u > 0), getting √ 1 − cos2 u . −2 = cos u Now square both sides of the equation above, then multiply both sides by cos2 u and rearrange to get the equation 5 cos2 u = 1. 1 1 Thus cos u = − √5 (the possibility that cos u equals √5 is eliminated π because 2 < u < π , which implies that cos u < 0). This can be written as cos u = − √ 5 5 . Student Solution Manual, Section 5.6 43. cos(−u) √ 5 solution cos(−u) = cos u = − 5 Exercise 43 Instructor’s Solutions Manual, Section 5.6 45. sin u solution sin u = 1 − cos2 u 1 = 1− 5 4 = 5 2 = √ 5 √ 2 5 = 5 Exercise 45 Student Solution Manual, Section 5.6 47. sin(−u) solution √ 2 5 sin(−u) = − sin u = − 5 Exercise 47 Instructor’s Solutions Manual, Section 5.6 49. cos(u + 4π ) solution √ 5 cos(u + 4π ) = cos u = − 5 Exercise 49 Student Solution Manual, Section 5.6 51. sin(u − 6π ) solution √ 2 5 sin(u − 6π ) = sin u = 5 Exercise 51 Instructor’s Solutions Manual, Section 5.6 53. tan(u + 8π ) solution tan(u + 8π ) = tan u = −2 Exercise 53 Student Solution Manual, Section 5.6 55. cos(u − 3π ) solution √ 5 cos(u − 3π ) = − cos u = 5 Exercise 55 Instructor’s Solutions Manual, Section 5.6 57. sin(u + 5π ) solution √ 2 5 sin(u + 5π ) = − sin u = − 5 Exercise 57 Student Solution Manual, Section 5.6 59. tan(u − 9π ) solution tan(u − 9π ) = tan u = −2 Exercise 59 Instructor’s Solutions Manual, Section 5.6 61. cos( π2 − u) solution π cos( 2 √ 2 5 − u) = sin u = 5 Exercise 61 Student Solution Manual, Section 5.6 Exercise 63 63. sin( π2 − u) solution sin π 2 √ 5 − u = cos u = − 5 Instructor’s Solutions Manual, Section 5.6 Exercise 65 65. tan( π2 − u) solution π tan( 2 − u) = 1 1 =− tan u 2 Student Solution Manual, Section 5.7 Exercise 1 Solutions to Exercises, Section 5.7 1. Evaluate cos−1 21 . solution cos π 3 = 12 ; thus cos−1 1 2 = π 3. Instructor’s Solutions Manual, Section 5.7 3. Evaluate tan−1 (−1). π π solution tan(− 4 ) = −1; thus tan−1 (−1) = − 4 . Exercise 3 Student Solution Manual, Section 5.7 Exercise 5 c Ν b u a Use the right triangle above for Exercises 5–12. This triangle is not drawn to scale corresponding to the data in the exercises. 5. Suppose a = 2 and c = 3. Evaluate u in radians. solution Because the cosine of an angle in a right triangle equals the length of the adjacent side divided by the length of the hypotenuse, we have cos u = 23 . Using a calculator working in radians, we then have u = cos−1 2 3 ≈ 0.841 radians. Instructor’s Solutions Manual, Section 5.7 7. Exercise 7 Suppose a = 2 and c = 5. Evaluate ν in radians. solution Because the sine of an angle in a right triangle equals the length of the opposite side divided by the length of the hypotenuse, we 2 have sin ν = 5 . Using a calculator working in radians, we then have ν = sin−1 2 5 ≈ 0.412 radians. Student Solution Manual, Section 5.7 9. Exercise 9 Suppose a = 5 and b = 4. Evaluate u in degrees. solution Because the tangent of an angle in a right triangle equals the length of the opposite side divided by the length of the adjacent side, we 4 have tan u = 5 . Using a calculator working in degrees, we then have u = tan−1 4 5 ≈ 38.7◦ . Instructor’s Solutions Manual, Section 5.7 11. Exercise 11 Suppose a = 5 and b = 7. Evaluate ν in degrees. solution Because the tangent of an angle in a right triangle equals the length of the opposite side divided by the length of the adjacent side, we 5 have tan ν = 7 . Using a calculator working in degrees, we then have ν = tan−1 5 7 ≈ 35.5◦ . Student Solution Manual, Section 5.7 13. Exercise 13 Find the smallest positive number t such that 10cos t = 6. solution The equation above implies that cos t = log 6. Thus we take t = cos−1 (log 6) ≈ 0.67908. Instructor’s Solutions Manual, Section 5.7 15. Exercise 15 Find the smallest positive number t such that etan t = 15. solution The equation above implies that tan t = ln 15. Thus we take t = tan−1 (ln 15) ≈ 1.21706. Student Solution Manual, Section 5.7 17. Exercise 17 Find the smallest positive number y such that cos(tan y) = 0.2. solution The equation above implies that we should choose tan y = cos−1 0.2 ≈ 1.36944. Thus we should choose y ≈ tan−1 1.36944 ≈ 0.94007. Instructor’s Solutions Manual, Section 5.7 19. Exercise 19 Find the smallest positive number x such that sin2 x − 3 sin x + 1 = 0. solution as Write y = sin x. Then the equation above can be rewritten y 2 − 3y + 1 = 0. Using the quadratic formula, we ﬁnd that the solutions to this equation are √ 3+ 5 ≈ 2.61803 y= 2 and √ 3− 5 ≈ 0.38197. y= 2 Thus sin x ≈ 2.61803 or sin x ≈ 0.381966. However, there is no real number x such that sin x ≈ 2.61803 (because sin x is at most 1 for every real number x), and thus we must have sin x ≈ 0.381966. Thus x ≈ sin−1 0.381966 ≈ 0.39192. Student Solution Manual, Section 5.7 21. Exercise 21 Find the smallest positive number x such that cos2 x − 0.5 cos x + 0.06 = 0. solution as Write y = cos x. Then the equation above can be rewritten y 2 − 0.5y + 0.06 = 0. Using the quadratic formula or factorization, we ﬁnd that the solutions to this equation are y = 0.2 and y = 0.3. Thus cos x = 0.2 or cos x = 0.3, which suggests that we choose x = cos−1 0.2 or x = cos−1 0.3. Because arccosine is a decreasing function, cos−1 0.3 is smaller than cos−1 0.2. Because we want to ﬁnd the smallest positive value of x satisfying the original equation, we choose x = cos−1 0.3 ≈ 1.2661. Instructor’s Solutions Manual, Section 5.8 Exercise 1 Solutions to Exercises, Section 5.8 1. Suppose t is such that cos−1 t = 2. Evaluate the following: (c) sin−1 (−t) (a) cos−1 (−t) −1 (b) sin t solution (a) cos−1 (−t) = π − cos−1 t = π − 2 (b) sin−1 t = −1 (c) sin π 2 − cos−1 t = −1 (−t) = − sin π 2 −2 t =2− π 2 Student Solution Manual, Section 5.8 Exercise 3 3. Suppose t is such that tan−1 t = (a) tan−1 3π 7 . Evaluate the following: (c) tan−1 (− t ) 1 1 t (b) tan−1 (−t) solution (a) Because t = tan 3π 7 , we see that t > 0. Thus tan−1 1 t = π 2 − tan−1 t = 3π (b) tan−1 (−t) = − tan−1 t = − 7 π (c) tan−1 (− 1t ) = − tan−1 ( 1t ) = − 14 π 2 − 3π 7 = π 14 . Instructor’s Solutions Manual, Section 5.8 Exercise 5 5. Evaluate cos(cos−1 41 ). 1 solution Let θ = cos−1 4 . Thus θ is the angle in [0, π ] such that 1 1 1 cos θ = 4 . Thus cos(cos−1 4 ) = cos θ = 4 . Student Solution Manual, Section 5.8 Exercise 7 7. Evaluate sin−1 (sin 2π 7 ). solution Let θ = sin−1 (sin π π interval [− 2 , 2 ] such that 2π 7 ). Thus θ is the unique angle in the sin θ = sin 1 2 1 Because − 2 ≤ 7 ≤ 2 , we see that 2π above implies that θ = 7 . 2π 7 2π 7 . π is in [− 2 , π 2 ]. Thus the equation Instructor’s Solutions Manual, Section 5.8 Exercise 9 9. Evaluate cos−1 (cos 3π ). solution Because cos 3π = −1, we see that cos−1 (cos 3π ) = cos−1 (−1). Because cos π = −1, we have cos−1 (−1) = π (cos 3π also equals −1, but cos−1 (−1) must be in the interval [0, π ]). Thus cos−1 (cos 3π ) = π . Student Solution Manual, Section 5.8 Exercise 11 11. Evaluate tan−1 (tan 11π 5 ). solution Because tan−1 is the inverse of tan, it may be tempting to think 11π 11π that tan−1 (tan 5 ) equals 5 . However, the values of tan−1 must be π π 11π π 11π between − 2 and 2 . Because 5 > 2 , we conclude that tan−1 (tan 5 ) 11π cannot equal 5 . Note that tan 11π 5 = tan(2π + Because π 5 π is in (− 2 , π 2 ), π 5) = tan π5 . we have tan−1 (tan tan−1 (tan 11π 5 ) π 5) = π 5. = tan−1 (tan π5 ) = Thus π 5. Instructor’s Solutions Manual, Section 5.8 13. Evaluate sin(− sin−1 Exercise 13 3 13 ). solution sin(− sin−1 3 13 ) = − sin(sin−1 3 = − 13 3 13 ) Student Solution Manual, Section 5.8 Exercise 15 15. Evaluate sin(cos−1 31 ). solution We give two ways to work this exercise: the algebraic approach and the right-triangle approach. 1 Algebraic approach: Let θ = cos−1 3 . Thus θ is the angle in [0, π ] such 1 that cos θ = 3 . Note that sin θ ≥ 0 because θ is in [0, π ]. Thus sin(cos−1 3 ) = sin θ = 1 − cos2 θ 1 = 1− 9 8 = 9 1 = √ 2 2 3 . 1 1 Right-triangle approach: Let θ = cos−1 3 ; thus cos θ = 3 . Because cos θ = adjacent side hypotenuse in a right triangle with an angle of θ, the following ﬁgure (which is not drawn to scale) illustrates the situation: Instructor’s Solutions Manual, Section 5.8 Exercise 15 3 Θ b 1 We need to evaluate sin θ. In terms of the ﬁgure above, we have sin θ = b opposite side = . hypotenuse 3 Applying the Pythagorean Theorem to the triangle above, we have b2 +1 = √ √ √ 2 2 9, which implies that b = 8 = 2 2. Thus sin θ = 3 . In other words, 1 sin(cos−1 3 ) = √ 2 2 3 . Student Solution Manual, Section 5.8 Exercise 17 17. Evaluate tan(cos−1 31 ). solution We give two ways to work this exercise: the algebraic approach and the right-triangle approach. Algebraic approach: From Exercise 15, we already know that sin(cos−1 3 ) = 1 √ 2 2 3 . Thus tan(cos−1 3 ) = 1 = sin(cos−1 31 ) 1 cos(cos−1 3 ) √ 2 2 3 1 3 √ = 2 2. 1 1 Right-triangle approach: Let θ = cos−1 3 ; thus cos θ = 3 . Because cos θ = adjacent side hypotenuse in a right triangle with an angle of θ, the following ﬁgure (which is not drawn to scale) illustrates the situation: Instructor’s Solutions Manual, Section 5.8 Exercise 17 3 Θ b 1 We need to evaluate tan θ. In terms of the ﬁgure above, we have tan θ = opposite side = b. adjacent side Applying the Pythagorean Theorem to the triangle above, we have b2 +1 = √ √ √ 9, which implies that b = 8 = 2 2. Thus tan θ = 2 2. In other words, √ 1 tan(cos−1 3 ) = 2 2. Student Solution Manual, Section 5.8 Exercise 19 19. Evaluate cos tan−1 (−4) . solution We give two ways to work this exercise: the algebraic approach and the right-triangle approach. π π Algebraic approach: Let θ = tan−1 (−4). Thus θ is the angle in (− 2 , 2 ) π π such that tan θ = −4. Note that cos θ > 0 because θ is in (− 2 , 2 ). Recall that dividing both sides of the identity cos2 θ + sin2 θ = 1 by cos2 θ 1 produces the equation 1 + tan2 θ = cos2 θ . Solving this equation for cos θ gives the following: cos tan−1 (−4) = cos θ 1 = 1 + tan2 θ 1 = 1 + (−4)2 = √1 17 = √ 17 17 . Right-triangle approach: Sides with a negative length make no sense in a right triangle. Thus ﬁrst we use some identities to get rid of the minus sign, as follows: cos tan−1 (−4) = cos(− tan−1 4) = cos(tan−1 4). Instructor’s Solutions Manual, Section 5.8 Exercise 19 Thus we need to evaluate cos(tan−1 4). Now let θ = tan−1 4; thus tan θ = 4. Because tan θ = opposite side adjacent side in a right triangle with an angle of θ, the following ﬁgure (which is not drawn to scale) illustrates the situation: c Θ 4 1 We need to evaluate cos θ. In terms of the ﬁgure above, we have cos θ = 1 adjacent side = . hypotenuse c 2 Applying the Pythagorean Theorem to the triangle above, we √ have c = √ 1 17 1 + 16, which implies that c = 17. Thus cos θ = √17 = 17 . In other √ √17 17 words, cos(tan−1 4) = 17 . Thus cos tan−1 (−4) = 17 . Student Solution Manual, Section 5.8 Exercise 21 21. Evaluate sin−1 (cos 2π 5 ). solution sin−1 (cos 2π 5 ) = π 2 − cos−1 (cos = π 2 − 2π 5 = π 10 2π 5 )

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