Right-Triangle-Based Definitions of the Trigonometric Functions Angles in standard position can be used to define the trigonometric functions.
There is also another way to approach them: as ratios of the lengths of the sides of right triangles.
Figure 27 shows an acute angle A in standard position. The definitions of the trigonometric func- tion values of angle A require x, y, and r. As drawn in Figure 27, x and y are the lengths of the two legs of the right triangle ABC, and r is the length of the hypotenuse.
The side of length y is the side opposite angle A, and the side of length x is the side adjacent to angle A. We use the lengths of these sides to replace x and y in the definitions of the trigonometric func-
tions, and the length of the hypotenuse to replace r, to obtain the following right- triangle-based definitions. In the definitions, we use the standard abbreviations for the sine, cosine, tangent, cosecant, secant, and cotangent functions.
5.3 Trigonometric Function Values and Angle Measures
■ Right-Triangle-Based Definitions of the Trigonometric Functions
■ Cofunctions
■ Trigonometric Function Values of Special Angles
■ Reference Angles
■ Special Angles as Reference Angles
■ Determination of Angle Measures with Special Reference Angles
■ Calculator Approximations of Trigonometric Function Values
■ Calculator
Approximations of Angle Measures
■ An Application
Right-Triangle-Based Definitions of Trigonometric Functions Let A represent any acute angle in standard position.
sin A∙ y
r ∙ side opposite A
hypotenuse csc A∙ r
y ∙ hypotenuse side opposite A cos A∙ x
r ∙ side adjacent to A
hypotenuse sec A∙ r
x ∙ hypotenuse side adjacent to A tan A∙ y
x ∙ side opposite A
side adjacent to A cot A∙ x
y ∙ side adjacent to A side opposite A
NOTE We will sometimes shorten wording like “side opposite A” to just
“side opposite” when the meaning is obvious.
Figure 27
A x
y
(x, y)
x C r y
B
EXAMPLE 1 Finding Trigonometric Function Values of an Acute Angle Find the sine, cosine, and tangent values for angles A and B in the right triangle in Figure 28.
SOLUTION The length of the side opposite angle A is 7, the length of the side adjacent to angle A is 24, and the length of the hypotenuse is 25.
sin A= side opposite hypotenuse = 7
25 cos A= side adjacent hypotenuse = 24
25 tan A= side opposite side adjacent = 7
24 The length of the side opposite angle B is 24, and the length of the side adjacent to angle B is 7.
sin B = 24
25 cos B = 7
25 tan B= 24 7
Use the right-triangle-based definitions of the trigonometric functions.
■✔ Now Try Exercise11.
Figure 28 A
B
7 24
25
C
The cofunction identities state the following.
Cofunction values of complementary angles are equal.
Cofunction Identities
For any acute angle A, the following hold.
sin A∙ cos190∙ ∙ A2 sec A∙ csc190∙ ∙ A2 tan A∙ cot190∙ ∙ A2 cos A ∙sin190∙ ∙ A2 csc A∙ sec190∙ ∙ A2 cot A ∙ tan190∙ ∙ A2
EXAMPLE 2 Writing Functions in Terms of Cofunctions Write each function in terms of its cofunction.
(a) cos 52° (b) tan 71° (c) sec 24°
SOLUTION
(a) Cofunctions
cos 52° =sin190°- 52°2 = sin 38° cos A=sin190° -A2 Complementary angles
(b) tan 71° = cot190° -71°2 = cot 19° (c) sec 24° = csc 66°
■✔ Now Try Exercises27 and 29.
Figure 29
A C
a c
b B
Whenever we use A, B, and C to name angles in a right triangle, C will be the right angle.
Whenever we use A, B, and C to name angles in a right triangle, C will be the right angle.
NOTE The cosecant, secant, and cotangent ratios are reciprocals of the sine, cosine, and tangent values, respectively, so in Example 1 we have
csc A= 25
7 sec A = 25
24 cot A= 24 7 csc B= 25
24 sec B = 25
7 and cot B = 7 24 .
Cofunctions Figure 29 shows a right triangle with acute angles A and B and a right angle at C. The length of the side opposite angle A is a, and the length of the side opposite angle B is b. The length of the hypotenuse is c. By the pre- ceding definitions, sin A = ac . Also, cos B =ac . Thus, we have the following.
sin A∙ a
c ∙ cos B Similarly, tan A∙ a
b ∙ cot B and sec A∙ c
b ∙csc B.
In any right triangle, the sum of the two acute angles is 90°, so they are complementary. In Figure 29, A and B are thus complementary, and we have established that sin A= cos B. This can also be written as follows.
sin A= cos190° - A2 B=90°-A
This is an example of a more general relationship between cofunction pairs.
sine, cosine tangent, cotangent secant, cosecant (+)+*
Cofunction pairs
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5.3 Trigonometric Function Values and Angle Measures
Trigonometric Function Values of Special Angles Certain special angles, such as 30°, 45°, and 60°, occur so often in trigonometry and in more advanced mathematics that they deserve special study. We start with an equilateral triangle, a triangle with all sides of equal length. Each angle of such a triangle measures 60°. Although the results we will obtain are independent of the length, for convenience we choose the length of each side to be 2 units. See Figure 30(a).
Bisecting one angle of this equilateral triangle leads to two right triangles, each of which has angles of 30°, 60°, and 90°, as shown in Figure 30(b). An angle bisector of an equilateral triangle also bisects the opposite side. Thus the shorter leg has length 1. Let x represent the length of the longer leg.
22 = 12 +x2 Pythagorean theorem 4= 1 +x2 Apply the exponents.
3= x2 Subtract 1 from each side.
23= x Square root property;
choose the positive root.
Figure 31 summarizes our results using a 30°9 60° right triangle. As shown in the figure, the side opposite the 30° angle has length 1. For the 30° angle,
hypotenuse= 2, side opposite= 1, side adjacent = 23 . Now we use the definitions of the trigonometric functions.
sin 30° = side opposite hypotenuse = 1
2 cos 30° = side adjacent
hypotenuse = 23 2 tan 30° = side opposite
side adjacent = 1 23 = 1
23 # 23
23 = 23 3 csc 30° = 2
1 = 2 sec 30° = 2
23 = 2
23 # 23
23 = 223 3 cot 30° = 23
1 = 23
Rationalize the denominators.
Figure 31 60° 2 30°
1
√3 60°
60°
60°
2
2 2
Equilateral triangle
(b) Figure 30 60°
30°
60° 1
2 2
30°
1 90° 90°
x x
30°– 60° right triangle (a)
EXAMPLE 3 Finding Trigonometric Function Values for 60∙
Find the six trigonometric function values for a 60° angle.
SOLUTION Refer to Figure 31 to find the following ratios.
sin 60° = 23
2 cos 60° = 1
2 tan 60° = 23 1 = 23 csc 60° = 2
23 = 223
3 sec 60° = 2
1 = 2 cot 60°= 1
23 = 23 3
■✔ Now Try Exercises 31, 33, and 35.
We find the values of the trigonometric functions for 45° by starting with a 45°9 45° right triangle, as shown in Figure 32. This triangle is isosceles. For simplicity, we choose the lengths of the equal sides to be 1 unit. (As before, the results are independent of the length of the equal sides.) If r represents the length of the hypotenuse, then we can find its value using the Pythagorean theorem.
12 + 12= r 2 Pythagorean theorem 2= r 2 Simplify.
22= r Choose the positive root.
Now we use the measures indicated on the 45°9 45° right triangle in Figure 32. sin 45° = 1
22 = 22
2 cos 45° = 1
22 = 22
2 tan 45° = 1 1 = 1 csc 45°= 22
1 = 22 sec 45° = 22
1 = 22 cot 45° = 1 1 = 1 Function values for 30°, 45°, and 60° are summarized in the table that follows.
NOTE The results in Example 3 can also be found using the fact that cofunction values of the complementary angles 60° and 30° are equal.
Function Values of Special Angles
U sin U cos U tan U cot U sec U csc U
30∙ 1
2 !3
2 !3
3 !3 2 !3
3 2
45∙ !2
2 !2
2 1 1 !2 !2
60∙ !3
2
1
2 !3 !3
3 2 2 !3
3
NOTE You will be able to reproduce this table quickly if you learn the values of sin 30°, sin 45°, and sin 60°. Then you can complete the rest of the table using the reciprocal, cofunction, and quotient identities.
Figure 32 45° 45°
1 r = √2 1
45°– 45° right triangle
Figure 33 on the next page shows several angles u (each less than one com- plete counterclockwise revolution) in quadrants II, III, and IV, respectively, with the reference angle u′ also shown. In quadrant I, angles u and u′ are the same. If an angle u is negative or has measure greater than 360°, its reference angle is found by first finding its coterminal angle that is between 0° and 360°, and then using the diagrams in Figure 33.
NOTE Reference angles are always positive and are between 0° and 90°.
Reference Angles Associated with every nonquadrantal angle in stand- ard position is an acute angle called its reference angle. A reference angle for an angle u, written u′, is the acute angle made by the terminal side of angle u and the x-axis.
551
5.3 Trigonometric Function Values and Angle Measures
CAUTION A common error is to find the reference angle by using the terminal side of u and the y-axis. The reference angle is always found with reference to the x-axis.
Figure 33 x
y
O
u in quadrant II u
u x
y
u in quadrant III O u
u
u in quadrant IV x y
O u
u
EXAMPLE 4 Finding Reference Angles Find the reference angle for each angle.
(a) 218° (b) 1387°
SOLUTION
(a) As shown in Figure 34(a), the positive acute angle made by the terminal side of this angle and the x-axis is
218° -180° = 38°.
For u= 218°, the reference angle u′ = 38°.
(b) First find a coterminal angle between 0° and 360°. Divide 1387° by 360° to obtain a quotient of about 3.9. Begin by subtracting 360° three times (because of the whole number 3 in 3.9).
1387°- 3 # 360°
= 1387° - 1080° Multiply.
= 307° Subtract.
The reference angle for 307° (and thus for 1387°) is 360° - 307° = 53°. See
Figure 34(b).
■✔ Now Try Exercises 57 and 61.
0 x
y
38°
218°
218° – 180° = 38°
(a) (b)
Figure 34
0 x
y
53°
307°
360° – 307° = 53°
*The authors would like to thank Bethany Vaughn and Theresa Matick, of Vincennes Lincoln High School, for their suggestions concerning this table.
EXAMPLE 5 Finding Trigonometric Function Values of a Quadrant III Angle
Find exact values of the six trigonometric functions of 210°.
SOLUTION An angle of 210° is shown in Figure 35. The reference angle is 210° -180° = 30°.
To find the trigonometric function values of 210°, choose point P on the ter- minal side of the angle so that the distance from the origin O to P is 2. (Any positive number would work, but 2 is most convenient.) By the results from 30°9 60° right triangles, the coordinates of point P become A-23 , -1B , with
x= -23 , y = -1, and r= 2. Then, by the definitions of the trigonometric functions, we obtain the following.
sin 210° = -1 2 = -1
2 csc 210° = 2 -1 = -2 cos 210° = -23
2 = - 23
2 sec 210°= 2
-23 = -223 3
Rationalize denominators as needed.
The preceding example suggests the following table for finding the refer- ence angle u′ for any angle u between 0° and 360°.
Special Angles as Reference Angles We can now find exact trigono- metric function values of angles with reference angles of 30°, 45°, or 60°.
Reference Angle U′ for U, where 0°*U*360° *
0 x
Q I y
0 x
Q II y
0 x
Q III y
0 x
Q IV y
u9 u9
u9 u9
u
u
u u
u9 = 180° – u
u9 = u – 180° u9 = 360° – u
u9 = u
Figure 35
x y
210°
30°
60°
O
P r x
y
y = –1 r = 2 x = –Ë3
tan 210° = -1
-23 = 23
3 cot 210° = -23 -1 = 23
■✔ Now Try Exercise 71.
553
5.3 Trigonometric Function Values and Angle Measures
Notice in Example 5 that the trigonometric function values of 210° cor- respond in absolute value to those of its reference angle 30°. The signs are different for the sine, cosine, secant, and cosecant functions because 210° is a quadrant III angle. These results suggest a shortcut for finding the trigonometric function values of a non-acute angle, using the reference angle.
In Example 5, the reference angle for 210° is 30°. Using the trigonometric function values of 30°, and choosing the correct signs for a quadrant III angle, we obtain the same results.
We determine the values of the trigonometric functions for any nonquadrantal angle u as follows. Keep in mind that all function values are positive when the terminal side is in Quadrant I, the sine and cosecant are positive in Quadrant II, the tangent and cotangent are positive in Quadrant III, and the cosine and secant are positive in Quadrant IV. In other cases, the function values are negative.
Finding Trigonometric Function Values for Any Nonquadrantal Angle Uθ
Step 1 If u7 360°, or if u6 0°, then find a coterminal angle by adding or subtracting 360° as many times as needed to obtain an angle greater than 0° but less than 360°.
Step 2 Find the reference angle u′.
Step 3 Find the trigonometric function values for reference angle u′.
Step 4 Determine the correct signs for the values found in Step 3. (Use the table of signs given earlier in the text or the paragraph above, if necessary.) This gives the values of the trigonometric functions for angle u.
NOTE To avoid sign errors when finding the trigonometric function val- ues of an angle, sketch it in standard position. Include a reference triangle complete with appropriate values for x, y, and r as done in Figure 35.
EXAMPLE 6 Finding Trigonometric Function Values Using Reference Angles
Find the exact value of each expression.
(a) cos1-240°2 (b) tan 675°
SOLUTION
(a) Because an angle of -240° is coterminal with an angle of -240° + 360° = 120°,
the reference angle is 180° - 120° = 60°, as shown in Figure 36. The cosine is negative in quadrant II.
cos1-240°2
= cos 120° Coterminal angle = -cos 60° Reference angle = -1
2 Evaluate.
Figure 36
x y
0 u = 60°
u = –240°
Figure 37 x y
0
u = 675° u = 45°
As shown in Figure 37, the reference angle is 360°- 315° =45°. An angle of 315° is in quadrant IV, so the tangent will be negative.
tan 675°
= tan 315° Coterminal angle
= - tan 45° Reference angle; quadrant-based sign choice
= -1 Evaluate. ■✔ Now Try Exercises 89 and 91.
(b) Subtract 360° to find an angle between 0° and 360° coterminal with 675°.
675° - 360°= 315°
EXAMPLE 7 Finding Angle Measures
Find all values of u, if u is in the interval 30°, 360°2 and cos u = -222 .
SOLUTION The value of cos u is negative, so u may lie in either quadrant II or III.
Because the absolute value of cos u is 222 , the reference angle u′ must be 45°. The two possible angles u are sketched in Figure 38.
180° - 45° = 135° Quadrant II angle u (from Figure 38 (a)) 180° + 45° = 225° Quadrant III angle u (from Figure 38 (b))
(a) (b)
Figure 38 x y
0 u = 45°
u in quadrant II u = 135°
x y
u in quadrant III 0
u = 225°
u = 45°
■✔ Now Try Exercise 97.
Determination of Angle Measures with Special Reference Angles The ideas discussed in this section can be used “in reverse” to find the measures of certain angles, given a trigonometric function value and an interval in which the angle must lie. We are most often interested in the interval 30°, 360°2.
CAUTION When evaluating trigonometric functions of angles given in degrees, the calculator must be in degree mode. An easy way to check this is to enter sin 90. The displayed answer should be 1. Also, if the angle or the reference angle is not a special or quadrantal angle, then the value given by the calculator is an approximation. And even if the angle or reference angle is a special angle, the value given by the calculator will often be an approximation.
Degree mode
A calculator can be used to find exact values such as cos 1-240°2 and tan 675°.
Calculator Approximations of Trigonometric Function Values We have found exact function values for special angles and for angles having special reference angles. Calculators provide approximations for function values of angles that do not satisfy these conditions.
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5.3 Trigonometric Function Values and Angle Measures
EXAMPLE 8 Finding Function Values with a Calculator Approximate the value of each expression.
(a) sin 49° 12′ (b) sec 97.977° (c) 1
cot 51.4283° (d) sin1-246°2 SOLUTION See Figure 39. We give values to eight decimal places below.
(a) We may begin by converting 49° 12′ to decimal degrees.
49° 12′ = 49 12 60
° =49.2°
However, some calculators allow direct entry of degrees, minutes, and seconds. (The method of entry varies among models.) Entering either sin149° 12′2 or sin 49.2° gives the same approximation.
sin 49° 12′= sin 49.2°≈ 0.75699506
(b) There are no dedicated calculator keys for the secant, cosecant, and cotan- gent functions. However, we can use reciprocal identities to evaluate them.
Recall that sec u= cos u1 for all angles u, where cos u∙0. Therefore, we use the reciprocal of the cosine function to evaluate the secant function.
sec 97.977° = 1
cos 97.977° ≈ -7.20587921
(c) Use the reciprocal identity cot u1 = tan u to simplify the expression first.
1
cot 51.4283° = tan 51.4283° ≈ 1.25394815
(d) sin1-246°2 ≈0.91354546 ■✔ Now Try Exercises 109, 111, 115, and 119.
Degree mode Figure 39
EXAMPLE 9 Using Inverse Trigonometric Functions to Find Angles Find an angle u in the interval 30°, 90°2 that satisfies each condition.
sin u= 0.96770915 (b) sec u =1.0545829 SOLUTION
(a) Using degree mode and the inverse sine function, we find that an angle u having sine value 0.96770915 is 75.399995°. (There are infinitely many such angles, but the calculator gives only this one.)
u= sin-1 0.96770915≈ 75.399995°
See Figure 40.
Degree mode Figure 40
Calculator Approximations of Angle Measures To find the measure of an angle having a certain trigonometric function value, calculators have three inverse functions (denoted sin∙1 , cos∙1 , and tan∙1).
If x is an appropriate number, then sin∙1 x, cos∙1 x, or tan∙1 x gives the measure of an angle whose sine, cosine, or tangent, respectively, is x.
For applications in this chapter, these functions will return angles in quadrant I.
CAUTION Compare Examples 8(b) and 9(b).
• To determine the secant of an angle, as in Example 8(b), we find the reciprocal of the cosine of the angle.
• To determine an angle with a given secant value, as in Example 9(b), we find the inverse cosine of the reciprocal of the value.
(b) Use the identity cos u= sec u1 . If sec u= 1.0545829 , then
cos u= 1
1.0545829 . Now, find u using the inverse cosine function.
u= cos-1 a 1
1.0545829b ≈18.514704° See Figure 40 on the previous page.
■✔ Now Try Exercises 125 and 129.
Figure 41 u > 0°
u < 0°
EXAMPLE 10 Finding Grade Resistance
When an automobile travels uphill or downhill on a highway, it experiences a force due to gravity. This force F in pounds is the grade resistance and is modeled by
F= W sin u,
where u is the grade and W is the weight of the automobile. If the automo- bile is moving uphill, then u7 0°; if downhill, then u 6 0°. See Figure 41. (Source: Mannering, F. and W. Kilareski, Principles of Highway Engineering and Traffic Analysis, Second Edition, John Wiley and Sons.)
An Application
(a) Calculate F to the nearest 10 lb for a 2500-lb car traveling an uphill grade with u= 2.5°.
(b) Calculate F to the nearest 10 lb for a 5000-lb truck traveling a downhill grade with u= -6.1°.
(c) Calculate F for u =0° and u =90°. Do these answers agree with intuition?
SOLUTION
(a) F= W sin u Given model for grade resistance F= 2500 sin 2.5° Substitute given values.
F≈ 110 lb Evaluate.
(b) F= W sin u= 5000 sin1-6.1°2≈ -530 lb F is negative because the truck is moving downhill.
(c) F= W sin u= W sin 0° =W102 = 0 lb F= W sin u= W sin 90° =W112 = W lb
This agrees with intuition because if u= 0°, then there is level ground and gravity does not cause the vehicle to roll. If u were 90°, the road would be vertical and the full weight of the vehicle would be pulled downward by gravity, so F = W.
■✔ Now Try Exercises 135 and 137.
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5.3 Trigonometric Function Values and Angle Measures