## Precalculus (6th Edition) Blitzer

a) The value is $-\frac{\sqrt{2}}{2}$. b) The value is $-\frac{\sqrt{2}}{2}$.
(a) In the unit circle, the point corresponding to $t=\frac{7\pi }{4}$ has the coordinates $\left( \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2} \right)$ And use $x=\frac{\sqrt{2}}{2}$ and $y=-\frac{\sqrt{2}}{2}$ such that, $\sin \left( \frac{7\pi }{4} \right)=y=-\frac{\sqrt{2}}{2}$ Thus, the value of the trigonometric function $\sin \left( \frac{7\pi }{4} \right)$ is, $-\frac{\sqrt{2}}{2}$. (b) The periodic properties of sine and cosine functions are, $\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$. Therefore, \begin{align} & \sin \left( \frac{47\pi }{4} \right)=\sin \left( \frac{7\pi }{4}+10\pi \right) \\ & =\sin \left( \frac{7\pi }{4}+5\left( 2\pi \right) \right) \\ & =\sin \frac{7\pi }{4} \end{align} Now put $\sin \frac{7\pi }{4}=-\frac{\sqrt{2}}{2}$. So, $\sin \left( \frac{7\pi }{4} \right)=-\frac{\sqrt{2}}{2}.$ Hence, the value of the trigonometric function $\sin \left( \frac{47\pi }{4} \right)$ is $-\frac{\sqrt{2}}{2}$.