Answer
a) The value is, $-\frac{\sqrt{2}}{2}$.
b)The value is, $-\frac{\sqrt{2}}{2}$.
Work Step by Step
(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, $-\frac{\sqrt{2}}{2}$.
