Answer
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 $\tan \frac{\pi }{4}=1$.
Work Step by Step
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 $\tan \frac{\pi }{4}=1$.
