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3. EVALUATION OF TRIGONOMETRIC FUNCTIONS In this section, we obtain values of the trigonometric functions for quadrantal angles, we introduce the idea of reference angles, and we discuss the use of a calculator to evaluate trigonometric functions of general angles. In Definition 2.1, the domain of each trigonometric function consists of all angles θ for which the denominator in the corresponding ratio is not zero. Because r > 0, it follows that sin θ = y / r and cos θ = x / r are defined for all angles θ . However, tan θ = y / x and sec θ = r / x are not defined when the terminal side of θ lies along the y axis (so that x = 0). Likewise, cot θ = x / y and csc θ = 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 angle, 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 + y2 = y y = =1 r y r y csc 90° = = =1 y y Therefore, sin 90° = 02 + y 2 = y 2 = y = y. 0 x =0 = r y x 0 cot 90° = = = 0. y y cos 90° = _______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 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 radians sin θ cos θ tan θ cot θ sec θ csc θ 0° 0 0 1 0 –– 1 –– 1 0 –– 0 –– 1 90° π 2 180° π 0 –1 0 –– –1 270° 3π 2 –1 0 –– 0 –– –1 360° 2π 0 1 0 –– 1 –– 17 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 cotermina; angle that is less than one revolution by adding or subtracting an integer multiple of 360° (or 2π radians). For instance, sin 450° = sin( 450° − 360° ) = sin 90° = 1. sec 7π = sec ( 7π − ( 3 × 2π ) ) = sec π = –1. cos (− 660°) = cos (− 660° + (2 × 360°) ) = cos 60° = 1 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 sin 1110° = sin 30° = csc 1110° = csc 30° = 2 2 3 2 3 sec 1110° = sec 30° = cos 1110° = cos 30° = 2 3 3 cot 1110° = cot 30° = 3 tan 1110° = tan 30° = 3 ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ Example 3.3 ---------------------------- -----------------------------------------------------------θ =– 5π 2 We repeatedly add 2 π to – – – 5π until we obtain a nonnegative coterminal angle: 2 5π π + 2π = – 2 2 π 2 + 2π = (still negative) 3π . 2 Therefore, by Table 3.1 for quadrantal angles, ⎛ 5π ⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ ⎛ 3π ⎞ cot ⎜ − ⎟ = cot ⎜ ⎟ = 0 sin ⎜ − ⎟ = sin ⎜ ⎟ = –1 2 2 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ 2 ⎠ ⎛ 5π ⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ ⎛ 3π ⎞ csc ⎜ − cos ⎜ − ⎟ = cos ⎜ ⎟ = 0 ⎟ = csc ⎜ ⎟ = –1 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 5π ⎞ ⎛ 5π ⎞ and both tan ⎜ − ⎟ and sec ⎜ − ⎟ are undefined. ⎝ 2 ⎠ ⎝ 2 ⎠ ______________________________________________________________________________________ 18 Table 3.2 θ degrees 30° θ radians π 6 π 45° 4 π 60° sin θ 1 2 cos θ tan θ cot θ sec θ csc θ 3 2 3 3 3 2 3 3 2 2 2 2 2 1 2 1 1 2 2 3 3 3 2 2 3 3 3 2 3 Figure 3.2 y y θ θR θ = θR O O x (a) y x (b) y θ R = θ ‐ 180° θ R = θ ‐ π θ θR θ R = 180° ‐ θ θ R = π ‐ θ θ R = 360° ‐ θ θ R = 2π ‐ θ θ O O x x θR (c) (d) 19 Example 3.4 ---------------------------- -----------------------------------------------------------Find the reference angle θ R for each angle θ . 3π (a) θ = 60° (b) θ = (c) θ = 210° 4 (a) By Figure 3.2(a), θ R = θ = 60° . (d) θ = 5π . 3 3π π = . 4 4 (c) By Figure 3.2(c), θ R = θ – 180° = 210° – 180° = 30° . π 5π (d) By Figure 3.2(c), θ R = 2 π – θ = 2 π – = . 3 3 ______________________________________________________________________________________ (b) By Figure 3.2(b), θ R = π – θ = π – 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, sin θ = ± sin θ R , cos θ = ± cos θ 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 define) 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] 2. (a) 5 π (b) 6 π (c) –7 π (d) 5π 2 (e) 7π . 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° 9. – 675° 11. 5 π 13. 17π 3 19π 2 25π 12. 6 31π 14. – 4 10. 20 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 θ R 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° 22. θ = 675° 21. θ = 315° 5π 6 13π =– 6 53π = 6 9π = 4 50π =– 3 147π =– 4 = – 5370° 23. θ = – 150° 24. θ = – 25. θ = – 60° 26. θ 27. θ = – 29. θ 31. θ 33. θ 35. θ π 4 2π =– 3 7π = 4 11π = 3 = – 420° 28. θ 30. θ 32. θ 34. θ 36. θ 37. Complete the following tables. (Do not use a calculator.) θ degrees 210° 225° 240° 300° 315° 330° θ radians sin θ cos θ 7π 6 5π 4 4π 3 5π 3 7π 4 11π 6 21 tan θ θ degrees 210° 225° 240° 300° 315° 330° θ radians cot θ sec θ csc θ 7π 6 5π 4 4π 3 5π 3 7π 4 11π 6 38. A calculator is set in radian mode. π is entered and the sine (SIN) key is pressed. The display shows – 4.1 × 10 -10 . But we know that sin π = 0. Explain. In problems 59 to 62, use a calculator to verify that the equation is true for the indicated value of the angle θ . 39. tan θ = sin θ cos θ for θ = 35° . 40. (cos θ )(tan θ ) = sin θ for θ = 41. cos 2 θ + sin 2 θ = 1 for θ = 5π 7 5π 3 42. 1 + tan 2 θ = sec 2 θ for θ = 17.75° 43. Verify that for θ = 0° , 30° , 45° , 60° , 90° , we have sin θ = 2 4 0 1 3 , , , , 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 sin θ .] 22