Answer
$\frac{1}{2}$
$0$
$-\frac{1}{2}$
$0$
$\frac{1}{2}$
Work Step by Step
Step 1. Given the function $y=f(x)=\frac{1}{2}cos(4x+\pi)$, we have $f(-\frac{\pi}{4})=\frac{1}{2}cos(4(-\frac{\pi}{4})+\pi)=\frac{1}{2}cos(0)=\frac{1}{2}$
Step 2. Similarly, we have $f(-\frac{\pi}{8})=\frac{1}{2}cos(4(-\frac{\pi}{8})+\pi)=\frac{1}{2}cos(\frac{\pi}{2})=0$
Step 3. We have $f(0)=\frac{1}{2}cos(4(0)+\pi)=\frac{1}{2}cos(\pi)=-\frac{1}{2}$
Step 4. We have $f(\frac{\pi}{8})=\frac{1}{2}cos(4(\frac{\pi}{8})+\pi)=\frac{1}{2}cos(\frac{3\pi}{2})=0$
Step 5. Finally, we have $f(\frac{\pi}{4})=\frac{1}{2}cos(4(\frac{\pi}{4})+\pi)=\frac{1}{2}cos(2\pi)=\frac{1}{2}$
