Answer
The exact value of the average rate of change of $f$ is $\frac{2\sqrt{2}-4}{\pi }$.
Work Step by Step
The slope of the line through the points $\left( {{x}_{1}},f\left( {{x}_{1}} \right) \right)$ and $\left( {{x}_{2}},f\left( {{x}_{2}} \right) \right)$ is defined as the average rate of change of the function.
Consider the following equation of slope $m$:
$m=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}}$
Put $\frac{3\pi }{2}$ for ${{x}_{2}}$ and $\frac{5\pi }{4}$ for ${{x}_{1}}$.
$m=\frac{f\left( \frac{3\pi }{2} \right)-f\left( \frac{5\pi }{4} \right)}{\frac{3\pi }{2}-\frac{5\pi }{4}}$
The condition is
$f\left( x \right)=\sin x$
Now, apply the condition in the above equation $m$.
$m=\frac{\sin \left( \frac{3\pi }{2} \right)-\sin \left( \frac{5\pi }{4} \right)}{\frac{3\pi }{2}-\frac{5\pi }{4}}$ …… (1)
The angle $\frac{5\pi }{4}$ lies between $\pi $ and $\frac{3\pi }{2}$, in quadrant III.
And, calculate the reference angle as follows
$\begin{align}
& {\theta }'=\frac{5\pi }{4}-\pi \\
& =\frac{5\pi -4\pi }{3} \\
& =\frac{\pi }{4}
\end{align}$
Hence,
$\sin \frac{5\pi }{4}=\frac{\sqrt{2}}{2}$
Here, the sine is negative in quadrant III.
So,
$\sin \frac{5\pi }{4}=-\sin \frac{\pi }{4}$
Put $\frac{\sqrt{2}}{2}$ for $\sin \frac{\pi }{4}$.
$\sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}$
Put $-1$ for $\sin \left( \frac{3\pi }{2} \right)$ and $-\frac{\sqrt{2}}{2}$ for $\sin \frac{5\pi }{4}$ in equation (1).
$\begin{align}
& m=\frac{\sin \left( \frac{3\pi }{2} \right)-\sin \left( \frac{5\pi }{4} \right)}{\frac{3\pi }{2}-\frac{5\pi }{4}} \\
& =\frac{\left( -1 \right)-\left( -\frac{\sqrt{2}}{2} \right)}{\frac{6\pi -5\pi }{4}} \\
& =\frac{-1+\frac{\sqrt{2}}{2}}{\frac{\pi }{4}}
\end{align}$
Further solve,
$\begin{align}
& m=\left( -1+\frac{\sqrt{2}}{2} \right)\frac{4}{\pi } \\
& =\frac{-4+\frac{\sqrt{2}}{2}\left( 4 \right)}{\pi } \\
& =\frac{2\sqrt{2}-4}{\pi }
\end{align}$
