## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 548: 56

#### Answer

a) The value is $1$. b) The value is $1$.

#### Work Step by Step

(a) In the provided unit circle, the point corresponding to $t=\frac{\pi }{2}$ has the coordinates $\left( 0,1 \right)$ Thus use $x=0$ and $y=1$ Such that, $\sin \left( \frac{\pi }{2} \right)=y=1$ Thus, the value of the trigonometric function $\sin \left( \frac{\pi }{2} \right)$ is $1$. (b) We know that the periodic properties of sine and cosine functions are, $\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$. So, \begin{align} & \sin \left( \frac{9\pi }{2} \right)=\sin \left( \frac{\pi }{2}+4\pi \right) \\ & =\sin \left( \frac{\pi }{2}+2\left( 2\pi \right) \right) \\ & =\sin \frac{\pi }{2} \end{align} Now, $\sin \left( \frac{\pi }{2} \right)=1$. So, $\sin \left( \frac{9\pi }{2} \right)=1$ Thus, the value of the trigonometric function $\sin \left( \frac{9\pi }{2} \right)$ is $1$.

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.