Answer
a) The value is $-\frac{\sqrt{2}}{2}$.
b) The value is $-\frac{\sqrt{2}}{2}$.
Work Step by Step
(a)
In the unit circle, the point corresponding to $t=\frac{7\pi }{4}$ has the coordinates $\left( \frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2} \right)$
And use $x=\frac{\sqrt{2}}{2}$ and $y=-\frac{\sqrt{2}}{2}$ such that,
$\sin \left( \frac{7\pi }{4} \right)=y=-\frac{\sqrt{2}}{2}$
Thus, the value of the trigonometric function $\sin \left( \frac{7\pi }{4} \right)$ is, $-\frac{\sqrt{2}}{2}$.
(b)
The periodic properties of sine and cosine functions are,
$\sin \left( t+2\pi \right)=\sin \left( t \right)$ and $\text{cos}\left( t+2\pi \right)=\cos \left( t \right)$.
Therefore,
$\begin{align}
& \sin \left( \frac{47\pi }{4} \right)=\sin \left( \frac{7\pi }{4}+10\pi \right) \\
& =\sin \left( \frac{7\pi }{4}+5\left( 2\pi \right) \right) \\
& =\sin \frac{7\pi }{4}
\end{align}$
Now put $\sin \frac{7\pi }{4}=-\frac{\sqrt{2}}{2}$. So,
$\sin \left( \frac{7\pi }{4} \right)=-\frac{\sqrt{2}}{2}.$
Hence, the value of the trigonometric function $\sin \left( \frac{47\pi }{4} \right)$ is $-\frac{\sqrt{2}}{2}$.
