## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 548: 57

#### Answer

a) The value is $0$. b) The value is $0$.

#### Work Step by Step

(a) As $\tan \left( t \right)=\frac{\sin \left( t \right)}{\cos \left( t \right)}$ So, $\tan \left( \pi \right)=\frac{\sin \left( \pi \right)}{\cos \left( \pi \right)}$ , In the unit circle, the point corresponding to $t=\frac{\pi }{2}$ has the coordinates $\left( -1,0 \right)$. Therefore, use $x=-1$ and $y=0$, such that, $\sin \left( \pi \right)=y=0\text{ }$ and $\cos \left( \pi \right)=x=-1$ So, \begin{align} & \tan \left( \pi \right)=\frac{\sin \left( \pi \right)}{\cos \left( \pi \right)} \\ & =\frac{0}{-1} \\ & =0 \end{align} Thus, the value of the trigonometric function $\tan \left( \pi \right)$ is $0$. (b) We know that the periodic properties of sine and cosine functions are, $\tan \left( t+\pi \right)=\tan \left( t \right)$ and $\text{cot}\left( t+\pi \right)=\cot \left( t \right)$. Therefore, \begin{align} & \tan \left( 17\pi \right)=\tan \left( \pi +16\pi \right) \\ & =\tan \left( \pi \right) \end{align} Now from $\tan \left( \pi \right)=0$. So, $\tan \left( 17\pi \right)=0$ Thus, the value of the trigonometric function $\tan \left( 17\pi \right)$ is $0$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.