## Precalculus (10th Edition)

Published by Pearson

# Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 87

#### Answer

$0$

#### Work Step by Step

I know that $\tan$ has a period of $180^\circ$, hence $\tan{\theta}=\tan{(\theta-180^\circ)}.$ Thus $\tan{200^\circ}=\tan{200^\circ-180^\circ}=\tan{20^\circ}$. I know that $\cos$ has a period of $360^\circ$, hence $\cos{\theta}=\cos{(\theta-360^\circ)}.$ Thus $\cos{380^\circ}=\cos{(380^\circ-360^\circ)}=\cos{20^\circ}.$ I know that $\sin$ is an even function, which means $f(-\theta)=-f(\theta).$ Therefore $\sin{-20^\circ}=-\sin{20^\circ}$. Thus: $\frac{\sin{-20^\circ}}{\cos{380^\circ}}+\tan{200^\circ}=\frac{-\sin{20^\circ}}{\cos{20^\circ}}+\tan{20^\circ}=-\tan{20^\circ}+\tan{20^\circ}=0$ Because $\frac{\sin{theta}}{\cos{theta}}=\tan{theta}$.

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