## Precalculus (6th Edition) Blitzer

a) The value is $0$. b) The value is $0$.
(a) The point corresponding to $t=\frac{\pi }{2}$ has the coordinates $\left( 1,0 \right)$ Thus, use $x=1$ and $y=0$ Such that, $\cos \left( \frac{\pi }{2} \right)=y=0$ Therefore, the value of the trigonometric function $\cos \left( \frac{\pi }{2} \right)$ is $0$. (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} & \cos \left( \frac{9\pi }{2} \right)=\cos \left( \frac{\pi }{2}+4\pi \right) \\ & =\cos \left( \frac{\pi }{2}+2\left( 2\pi \right) \right) \\ & =\cos \frac{\pi }{2} \end{align} Now, put $\cos \left( \frac{\pi }{2} \right)=0$. So, $\cos \left( \frac{9\pi }{2} \right)=0$ Thus, the value of the trigonometric function $\cos \left( \frac{9\pi }{2} \right)$ is $0$.