Trigonometry Teacher`s Edition - Solution Key - CK

Trigonometry Teacher`s Edition - Solution Key - CK
Trigonometry Teacher’s Edition - Solution Key
CK-12 Foundation
December 9, 2009
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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
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