## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 548: 58

#### Answer

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

#### Work Step by Step

(a) It is given that $\cot \left( t \right)=\frac{\cos \left( t \right)}{\sin \left( t \right)}$ So, $\cot \left( \frac{\pi }{2} \right)=\frac{\cos \left( \frac{\pi }{2} \right)}{\sin \left( \frac{\pi }{2} \right)}$ In the unit circle, the point corresponding to $t=\frac{\pi }{2}$ has the coordinates $\left( 0,1 \right)$. Therefore, use $x=0$ and $y=1$ such that, $\sin \left( \frac{\pi }{2} \right)=y=1$ and $\cos \left( \frac{\pi }{2} \right)=x=0$ So, \begin{align} & \cot \left( \frac{\pi }{2} \right)=\frac{\cos \left( \frac{\pi }{2} \right)}{\sin \left( \frac{\pi }{2} \right)} \\ & =\frac{0}{1} \\ & =0 \end{align} Thus, the value of the trigonometric function 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 $\text{cot}\left( t+\pi \right)=\cot \left( t \right)$. So, \begin{align} & \cot \left( \frac{15\pi }{2} \right)=\cot \left( \frac{\pi }{2}+7\pi \right) \\ & =\cot \left( \frac{\pi }{2} \right) \end{align} Now put $\cot \left( \frac{\pi }{2} \right)=0$. So, $\cot \left( \frac{15\pi }{2} \right)=0$ Thus, the value of the trigonometric function $\cot \left( \frac{15\pi }{2} \right)$ is $0$.

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