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..
HINT
From $\cos(2x) = 2 \cos^2(x) - 1$, replace each term already squared by something which looks that the cosine of the double angle. Since your base angle is $\pi/8$, the double angle is $\pi/4$ and you know the value of the cosine of this angle.
I am sure you can take from here.
For any $\alpha \in \mathbb{R}$, let $\lambda_k = \cos\left[\frac{\pi}{8}(2k-\alpha)\right]$ for $k = 1,2,\ldots,8$. All of them satisfy
$$\cos(8\cos^{-1}\lambda_k) = \cos\left[\pi(2k-\alpha)\right] = \cos\pi\alpha$$ This implies they are the 8 roots of a polynomial of degree 8:
$$T_8(x) -\cos\pi\alpha = 0 \quad\iff\quad 128x^8-256x^6+160x^4-32x^2 + (1-\cos\pi\alpha) = 0 $$
where $$T_8(x) = \cos(8\cos^{-1}x) = 128x^8-256x^6+160x^4-32x^2 + 1$$ is the Chebyshev polynomial of $1^{st}$ kind with degree $8$. Notice $\lambda_{4+k} = -\lambda_k$ for $k = 1,\ldots, 4$, this gives us
$$\begin{align} \prod_{k=1}^4(x - \lambda_k^2) = \prod_{k=1}^8(\sqrt{x}-\lambda_k) &= \frac{1}{128}(T_8(\sqrt{x})-\cos\pi\alpha)\\ &= x^4 - 2 x^3+\frac54 x^2-\frac14 x + \frac{1-\cos\pi\alpha}{128} \end{align}$$ By comparing the coefficients of $x^3$ and $x^2$, we get
$$\sum_{k=1}^4 \lambda_k^2 = 2\quad\text{ and }\quad \sum_{1\le k < \ell \le 4} \lambda_k^2\lambda_\ell^2 = \frac54$$
When $\alpha = 1$, $\;\displaystyle\sum_{k=1}^4 \lambda_k^4\;$ reduces to $$\cos^4\frac{\pi}{8} + \cos^4\frac{3\pi}{8}+\cos^4\frac{5\pi}{8}+\cos^4\frac{7\pi}{8}$$ and hence the desired sum can be evaluated as $$\sum_{k=1}^4 \lambda_k^4 = \left(\sum_{k=1}^4 \lambda_k^2 \right)^2 - 2 \left(\sum_{1\le k < \ell \le 4} \lambda_k^2\lambda_\ell^2\right) = 2^2 - 2\cdot\frac54 = \frac32$$
$\cos^4\frac{3\pi}{8}=\left(\frac{1+\cos \frac{3\pi}{4}}{2} \right)^2=(\frac{1-\sqrt{2}/2}{2})^2=(\frac{2-\sqrt{2}}{4})^2=\frac{6-4\sqrt{2}}{16}=\frac{3-2\sqrt{2}}{8}$
$\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} $
