## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# APPENDIX D - Trigonometry - D Exercises: 30

#### Answer

$sin \alpha=\frac{2\sqrt 5}{5}$ $cos \alpha=\frac{\sqrt 5}{5}$ $csc \alpha=\frac{\sqrt 5}{2}$ $sec \alpha=\sqrt 5$ $cot \alpha=\frac{1}{2}$

#### Work Step by Step

Since $tan \alpha=2$, we can label the opposite side as having length 2 and the adjacent side as having length 1. The Pythagorean Theorem gives hypotenuse side $=\sqrt {2^{2}+1^{2}}=\sqrt 5$ The other five trigonometric functions are given as follows: $sin \alpha=\frac{2\sqrt 5}{5}$ $cos \alpha=\frac{\sqrt 5}{5}$ $csc \alpha=\frac{\sqrt 5}{2}$ $sec \alpha=\sqrt 5$ $cot \alpha=\frac{1}{2}$

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