## Precalculus (6th Edition) Blitzer

a) The value is, $-\frac{\sqrt{2}}{2}$. b)The value is, $-\frac{\sqrt{2}}{2}$.
(a) In the given unit circle, the point corresponding to $t=\frac{3\pi }{4}$ has the coordinates $\left( -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$ Thus, use $x=-\frac{\sqrt{2}}{2}$ and $y=\frac{\sqrt{2}}{2}$ Such that, $\cos \left( \frac{3\pi }{4} \right)=x=-\frac{\sqrt{2}}{2}$ And the value of the trigonometric function $\cos \left( \frac{3\pi }{4} \right)$ is, $-\frac{\sqrt{2}}{2}$. (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{11\pi }{4} \right)=\cos \left( \frac{3\pi }{4}+2\pi \right) \\ & =\cos \left( \frac{3\pi }{4} \right) \end{align} Now put $\cos \left( \frac{13\pi }{4} \right)=-\frac{\sqrt{2}}{2}.$ $\cos \left( \frac{11\pi }{4} \right)=-\frac{\sqrt{2}}{2}$ Thus, the value of the trigonometric function $\cos \left( \frac{11\pi }{4} \right)$ is, $-\frac{\sqrt{2}}{2}$.