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 that corresponds to $t=\frac{3\pi }{4}$ has the coordinates $\left( -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$
So use $x=-\frac{\sqrt{2}}{2}$ and $y=\frac{\sqrt{2}}{2}$.
Such that,
$\sin \left( \frac{3\pi }{4} \right)=y=\frac{\sqrt{2}}{2}$
The value of the trigonometric function $\sin \left( \frac{3\pi }{4} \right)$ is $\frac{\sqrt{2}}{2}$.
(b)
We have the periodic properties of sine and cosine functions:
$\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$.
Therefore,
$\begin{align}
& \sin \left( \frac{11\pi }{4} \right)=\sin \left( \frac{3\pi }{4}+2\pi \right) \\
& =\sin \left( \frac{3\pi }{4} \right)
\end{align}$
Now, $\sin \left( \frac{3\pi }{4} \right)=\frac{\sqrt{2}}{2}$. So,
$\sin \left( \frac{3\pi }{4} \right)=\frac{\sqrt{2}}{2}.$
Thus, the value of the trigonometric function $\sin \left( \frac{11\pi }{4} \right)$ is, $\frac{\sqrt{2}}{2}$.
