## Precalculus (6th Edition) Blitzer

The exact value of the average rate of change of $f$ is $\frac{2\sqrt{2}-4}{\pi }$.
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}