The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$ Problem : 
The value of $\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$
If this could have been like this $\cos\frac{\pi}{8} + \cos\frac{3\pi}{8}+\cos\frac{5\pi}{8}+\cos\frac{7\pi}{8}$ then we can take the terms like 
$(\cos\frac{3\pi}{8}+\cos\frac{5\pi}{8} ) + (\cos\frac{\pi}{8} +\cos\frac{7\pi}{8})$
and solve further , but in this case due to power of 4 I am unable to proceed please suggest... thanks..

$\displaystyle \cos4x=2\cos^22x-1=2(2\cos^2x-1)^2-1=8\cos^4x-8\cos^2x+1 $
If $\displaystyle \cos4x=0, 4x=(2n+1)\frac\pi2$ where $n$ is any integer
So, the roots of $\displaystyle 8c^4-8c^2+1=0$ are $\displaystyle\cos(2n+1)\frac\pi8$ where $n=0,1,2,3$
If we set $\displaystyle c^4=d,$  we have $8d+1=8c^2\implies (8d+1)^2=(8c^2)^2=64d\iff 64d^2-48d+1=0\  \ \ \ (1)$
As $\displaystyle -\cos\frac{r\pi}8=\cos\left(\pi-\frac{r\pi}8\right)=\cos\frac{(8-r)\pi}8, \cos^2\frac{r\pi}8=\cos^2\frac{(8-r)\pi}8$
$\displaystyle\implies \sum_{n=0}^3\cos^4(2n+1)\frac\pi8=2 \sum_{n=0}^1\cos^4(2n+1)\frac\pi8$
Using Vieta's formula on $(1),$
$\displaystyle\sum_{n=0}^1\cos^4(2n+1)\frac\pi8=\frac34$

We can in fact, square $(1)$ to get $$64^2d^4-2\cdot64\cdot48d^2+\cdots+1=0$$
$\displaystyle\implies \sum_{n=0}^3\cos^4(2n+1)\frac\pi8=\frac{2\cdot64\cdot48}{64^2} $
