## Precalculus (6th Edition) Blitzer

The point $P$ on the unit circle that corresponds to $t=\frac{\pi }{4}$ has coordinates $\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ . Thus, $\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ , $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$, and $\tan \frac{\pi }{4}=1$.
Consider the provided point. $\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ Consider the provided angle. $t=\frac{\pi }{4}$ Consider the expression for $\sin t$. $\sin t=y$ Substitute $\frac{\sqrt{2}}{2}$ for $y$ and $\frac{\pi }{4}$ for $t$. $\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Consider the expression for $\cos t$. $\cos t=x$ Substitute $\frac{\sqrt{2}}{2}$ for $x$ and $\frac{\pi }{4}$ for $t$. $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ Consider the expression for $\tan t$. $\tan t=\frac{y}{x},x\ne 0$ Substitute $\frac{\sqrt{2}}{2}$ for $x$, $\frac{\sqrt{2}}{2}$ for $y$ and $\frac{\pi }{4}$ for $t$ : \begin{align} & \tan \frac{\pi }{4}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \\ & =1 \end{align} The point $P$ on the unit circle that corresponds to $t=\frac{\pi }{4}$ has coordinates $\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ . Thus, $\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}$ , $\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}$, and $\tan \frac{\pi }{4}=1$.