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F. Find the exact value of the following functions (using the procedure from the examples) |(a) sin600° (d) csc 35 (e) tan (b) tan (-810°) 11 (c) cos

F. Find the exact value of the following functions (using the procedure from the examples) |(a) sin600° (d) csc 35 (e) tan (b) tan (-810°) 11 (c) cos

Trigonometry (11th Edition)
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Hi, please help me solve Problem F ONLY using the examples given as a reference. Thank you!

F. Find the exact value of the following functions (using the procedure from the examples) |(a) sin600° (d) csc 35z (b) tan (-810°) (e) tan- 11m (c) cos
Transcribed Image Text:F. Find the exact value of the following functions (using the procedure from the examples) |(a) sin600° (d) csc 35z (b) tan (-810°) (e) tan- 11m (c) cos
•Trigonometric Functions of angles greater than 90° Note: The function of an angle greater than 90° is equal to (+) or (-) the function of its reference angle. The sign depends on the quadrant where the terminal side of 0 lies. Example: Find the exact value of the following trigonometric function. 1. tan150° 4. csc2100 2. sin240° 5. sec330° 3. tan315° Solution: 1. We find first for the reference angle of 0 = 150° e QII. 0 = 180° – 150° = 30° Note: Tangent in QIII is negative (-). 1 . tan 150° = - tan30° = V3 2. Reference angle 0 240°- 180° = 60° Note: Sine in QII is negative (-). V3 . sin 240° = -sin 60°= 2 By doing the same procedure from examples 1 and 2 above, we have the following results for the rest of the problems. 3. tan 315° = - tan45° = -1 4. csc 210° = - csc30° = -2 V3 5. sec 330° = - tan30° Note: Same procedure is applied when the angle measure is in radian. Illustration: Find the exact value of cos . Solution: The given angle in radian measure is more than 2 rotations. Take out the number of rotations in excess of one rotation then use the procedure above. Hence, 19n 3n 4n = v2 19n Thus, cos = cos = -cos 2
Transcribed Image Text:•Trigonometric Functions of angles greater than 90° Note: The function of an angle greater than 90° is equal to (+) or (-) the function of its reference angle. The sign depends on the quadrant where the terminal side of 0 lies. Example: Find the exact value of the following trigonometric function. 1. tan150° 4. csc2100 2. sin240° 5. sec330° 3. tan315° Solution: 1. We find first for the reference angle of 0 = 150° e QII. 0 = 180° – 150° = 30° Note: Tangent in QIII is negative (-). 1 . tan 150° = - tan30° = V3 2. Reference angle 0 240°- 180° = 60° Note: Sine in QII is negative (-). V3 . sin 240° = -sin 60°= 2 By doing the same procedure from examples 1 and 2 above, we have the following results for the rest of the problems. 3. tan 315° = - tan45° = -1 4. csc 210° = - csc30° = -2 V3 5. sec 330° = - tan30° Note: Same procedure is applied when the angle measure is in radian. Illustration: Find the exact value of cos . Solution: The given angle in radian measure is more than 2 rotations. Take out the number of rotations in excess of one rotation then use the procedure above. Hence, 19n 3n 4n = v2 19n Thus, cos = cos = -cos 2
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