Precalculus (6th Edition) Blitzer

The way to use the unit circle to find the value of the trigonometric function at $\frac{\pi }{4}$ is to find the points on the unit circle that corresponds to $t=\frac{\pi }{4}$.
First, find the points on the unit circle $\left( a,b \right)$ that corresponds to$t=\frac{\pi }{4}$. Consider the unit circle as shown in the figure: The point $P$ lies on the line $y=x$ Consider the standard equation of the circle${{x}^{2}}+{{y}^{2}}=1$ . Since, $y=x$ Substitute $y$ for $x$ in above equation to get: \begin{align} & {{y}^{2}}+{{y}^{2}}=1 \\ & 2{{y}^{2}}=1 \\ & y=\frac{1}{\sqrt{2}} \\ & y=\frac{\sqrt{2}}{2} \end{align} Thus, the point is $\left( \frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ Substitute these values in the formula of the trigonometric functions to get the required values.