Answer: \(\cos8\theta=128\cos^8\theta-256\cos^6\theta+160\cos^4\theta-32\cos^2\theta+1.\)

If \(x=\cos^2\frac{\pi}{8}\), then

\(4x^4-8x^3+5x^2-x=-\frac1{16}.\)

By de Moivre's theorem,

\((\cos\theta+i\sin\theta)^8=\cos8\theta+i\sin8\theta.\)

Taking the real part after expanding gives

\(\cos8\theta=\cos^8\theta-28\cos^6\theta\sin^2\theta+70\cos^4\theta\sin^4\theta-28\cos^2\theta\sin^6\theta+\sin^8\theta.\)

Replacing powers of \(\sin^2\theta\) by \(1-\cos^2\theta\) and simplifying gives

\(\cos8\theta=128\cos^8\theta-256\cos^6\theta+160\cos^4\theta-32\cos^2\theta+1.\)

Let

\(x=\cos^2\frac{\pi}{8}.\)

Putting \(\theta=\frac{\pi}{8}\) gives \(\cos8\theta=\cos\pi=-1\). Hence

\(-1=128x^4-256x^3+160x^2-32x+1.\)

So

\(128x^4-256x^3+160x^2-32x=-2.\)

Divide by \(32\):

\(4x^4-8x^3+5x^2-x=-\frac1{16}.\)