Answer
a) The value is $1$.
b) The value is $1$.
Work Step by Step
(a)
In the provided unit circle, the point corresponding to $t=\frac{\pi }{2}$ has the coordinates $\left( 0,1 \right)$
Thus use $x=0$ and $y=1$
Such that,
$\sin \left( \frac{\pi }{2} \right)=y=1$
Thus, the value of the trigonometric function $\sin \left( \frac{\pi }{2} \right)$ is $1$.
(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,
$\begin{align}
& \sin \left( \frac{9\pi }{2} \right)=\sin \left( \frac{\pi }{2}+4\pi \right) \\
& =\sin \left( \frac{\pi }{2}+2\left( 2\pi \right) \right) \\
& =\sin \frac{\pi }{2}
\end{align}$
Now, $\sin \left( \frac{\pi }{2} \right)=1$. So,
$\sin \left( \frac{9\pi }{2} \right)=1$
Thus, the value of the trigonometric function $\sin \left( \frac{9\pi }{2} \right)$ is $1$.
