<<

3. EVALUATION OF In this section, we obtain values of the trigonometric functions for quadrantal , we introduce the idea of reference angles, and we discuss the use of a to evaluate trigonometric functions of general angles. In Definition 2.1, the domain of each trigonometric consists of all angles θ for which the denominator in the corresponding ratio is not zero. Because r > 0, it follows that sinθ = y/r and cosθ = x/r are defined for all angles θ . However, tanθ = y/x and secθ = r/x are not defined when the terminal side of θ lies anlong the y axis (so that x = 0). Likewise, cotθ = x/y and cscθ = r/y are not defined when the terminal side of θ lies along the x axis (so that y = 0). Therefore, when you deal with a trigonometric function of a quadrantal , you must check to be sure that the function is actually defined for that angle.

Example 3.1 ------Find the values (if they are defined) of the six trigonometric functions for the quadrantal angle θ = 90° (or θ = π 2 ). In order to use Definition 1, we begin by choosing any point ( 0 , y ) with y > 0, on the terminal side of the 90° angle (Figure 1). Because x = 0, it follows that tan 90° and sec 90° are undefined. Since y > 0, we have r = x2 + y 2 = 02 + y 2 = y 2 = y = y. y y x 0 Therefore, sin 90° = = = 1 cos 90° = = = 0 r y r y r y x 0 csc90° = = = 1 cot 90° = = = 0. y y y y ______

The values of the trigonometric functions for other quadrantal angles are found in a similar manner. The results appear in Table 3.1. Dashes in the table indicate that the function is undefined for that angle.

Table 3.1

θ degrees θ sinθ cosθ tanθ cotθ secθ cscθ

0° 0 0 1 0 –– 1 ––

π 90° 1 0 –– 0 –– 1 2

180° π 0 –1 0 –– –1 ––

3π 270° –1 0 –– 0 –– –1 2

360° 2π 0 1 0 –– 1 ––

18

It follows from Definition 2.1 that the values of each of the six trigonometric functions remain unchanged if the angle is replaced by a coterminal angle. If an angle exceeds one revolution or is negative, you can change it to a nonnegative coterminal angle that is less than one revolution by adding or subtracting an multiple of 360° (or 2π radians). For instance, sin 450° = sin( 450° − 360° ) = sin 90° = 1.

sec 7π = sec () 7π − () 3 × 2π = secπ = –1. 1 cos ()− 660° = cos ()− 660° + (2 × 360°) = cos 60° = . 2

In Examples 3.2 and 3.3, replace each angle by a nonnegative coterminal angle that is less than on revolution and then find the values of the six trigonometric functions (if they are defined).

Example 3.2 ------θ = 1110° By dividing 1110 by 360, we find that the largest integer multiple of 360° that is less than 1110° is 3 × 360° = 1080° . Thus, 1110° – ( 3 × 360° ) = 1110° – 1080° = 30° . (Or we could have started with 1110° and repeatedly subtracted 360° until we obtained 30° .) It follows that 1 sin1110° = sin 30° = csc1110° = csc 30° = 2 2 3 2 3 cos1110° = cos 30° = sec1110° = sec 30° = 2 3 3 tan1110° = tan 30° = cot1110° = cot 30° = 3 3 ______

Example 3.3 ------5π θ = – 2 5π We repeatedly add 2π to – until we obtain a nonnegative coterminal angle: 2 5π π – + 2π = – (still negative) 2 2 π 3π – + 2π = . 2 2 Therefore, by Table 3.1 for quadrantal angles, ⎛ 5π ⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ ⎛ 3π ⎞ sin ⎜− ⎟ = sin ⎜ ⎟ = –1 cot ⎜− ⎟ = cot ⎜ ⎟ = 0 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 5π ⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ ⎛ 3π ⎞ cos ⎜− ⎟ = cos ⎜ ⎟ = 0 csc ⎜− ⎟ = csc ⎜ ⎟ = –1 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 5π ⎞ ⎛ 5π ⎞ and both tan ⎜− ⎟ and sec ⎜− ⎟ are undefined. ⎝ 2 ⎠ ⎝ 2 ⎠ ______

19

Table 3.1 θ degrees θ radians sinθ cosθ tanθ cotθ secθ cscθ π 1 3 3 2 3 30° 3 2 6 2 2 3 3 π 2 2 45° 1 1 2 2 4 2 2 π 3 1 3 2 3 60° 3 2 3 2 2 3 3

Figure 3.2

y y θ = 180° ‐ θ R θ R = π ‐ θ

θ θ R θ = θR

O O x x

(a) (b)

y y

θR = θ ‐ 180° θR = 360° ‐ θ

θ θ π θ π θ R = ‐ R = 2 ‐

θ θ

O x O x θR

θ R

(c) (d)

20

Example 3.4 ------

Find the reference angle θ R for each angle θ . 3π 5π (a) θ = 60° (b) θ = (c) θ = 210° (d) θ = . 4 3

(a) By Figure 3.2(a), θ R = θ = 60° . 3π π (b) By Figure 3.2(b), θ = π – θ = π – = . R 4 4

(c) By Figure 3.2(c), θ R = θ – 180° = 210° – 180° = 30° . 5π π (d) By Figure 3.2(c), θ = 2π – θ = 2π – = . R 3 3 ______

The value of any trigonometric function of any angle θ is the same as the value of the

function for the reference angle, θ R , except possibly for a change of algebraic sign.

Thus,

sinθ = ± sinθ R , cosθ = ± cosθ R , and so forth. You can always determine the correct algebraic sign by considering the quadrant in which θ lies.

Section 3 Problems------

In problems 1 and 2, find the values (if they are defined) of the six trigonometric functions of the given quadrantal angles. (Do not use a calculator.)

1. (a) 0° (b) 180° (c) 270° (d) 360° .

[When you have finished, compare your answers with the results in Table 3.1] 5π 7π 2. (a) 5π (b) 6π (c) –7π (d) (e) . 2 2

In Problems 3 to 14, replace each angle by a nonnegative coterminal angle that is less than one revolution and then find the exact values of the six trigonometric functions (if they are defined) for the angle. 3. 1440° 4. 810° 5. 900° 6. – 220° 7. 750° 8. 1845° 19π 9. – 675° 10. 2 25π 11. 5π 12. 6 13. 17π 14. – 31π 3 4

21

15. What happens when you try to evaluate tan 900° on a calculator? [Try it.]

16. Let θ be a quadrant III angle in standard position and let θ R be its reference angle. Show that the value of any trigonometric function of θ is the same as the value of

θ R , except possibly for a change of algebraic sign. Repeat for θ in quadrant IV.

In problems 17 to 36, find the reference angle θR for each angle θ , and then find the exact values of the six trigonometric functions of θ . 17. θ = 150° 18. θ = 120° 19. θ = 240° 20. θ = 225° 21. θ = 315° 22. θ = 675° 5π 23. θ = –150° 24. θ = – 6 13π 25. θ = – 60° 26. θ = – 6 π 53π 27. θ = – 28. θ = 4 6 2π 9π 29. θ = – 30. θ = 3 4 7π 50π 31. θ = 32. θ = – 4 3 11π 147π 33. θ = 34. θ = – 3 4 35. θ = – 420° 36. θ = – 5370°

37. Complete the following tables. (Do not use a calculator.)

θ degrees θ radians sinθ cosθ tanθ 7π 210° 6 5π 225° 4 4π 240° 3 5π 300° 3 7π 315° 4 11π 330° 6

22

θ degrees θ radians cotθ secθ cscθ 7π 210° 6 5π 225° 4 4π 240° 3 5π 300° 3 7π 315° 4 11π 330° 6

38. A calculator is set in mode. π is entered and the (SIN) key is pressed. The display shows – 4.1 × 10-10 . But we know that sinπ = 0. Explain.

In problems 59 to 62, use a calculator to verify that the is true for the indicated value of the angle θ . sinθ 59. tanθ = for θ = 35° . cosθ 5π 60. (cosθ )(tanθ ) = sinθ for θ = 7 5π 61. cos2 θ + sin 2 θ = 1 for θ = 3 62. 1+ tan 2 θ = sec2 θ for θ = 17.75°

0 1 2 3 4 63. Verify that for θ = 0° , 30° , 45° , 60° , 90° , we have sinθ = , , , , and 2 2 2 2 2 respectively. [Although there is no theoretical significance to this pattern, people often use it as a memory aid to help recall these values of sinθ .]

23