Prove that \(\cos^6\theta=\frac{1}{32}(\cos6\theta+6\cos4\theta+15\cos2\theta+10)\).

Let \(x=\cos\theta+i\sin\theta\). Then \(x+\frac1x=2\cos\theta\) and \(x^n+\frac1{x^n}=2\cos n\theta\).

\((2\cos\theta)^6=\left(x+\frac1x\right)^6=x^6+6x^4+15x^2+20+\frac{15}{x^2}+\frac6{x^4}+\frac1{x^6}\) (binomial coefficients 1, 6, 15, 20, 15, 6, 1)

\(=\left(x^6+\frac1{x^6}\right)+6\left(x^4+\frac1{x^4}\right)+15\left(x^2+\frac1{x^2}\right)+20\)

\(64\cos^6\theta=2\cos6\theta+12\cos4\theta+30\cos2\theta+20\)

\(\cos^6\theta=\frac1{32}\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)\) (proved). Check at \(\theta=0\): \(\frac{1+6+15+10}{32}=1\).