Answer: (a) \(\cos^4\theta=\frac{1}{8}\cos 4\theta+\frac{1}{2}\cos 2\theta+\frac{3}{8}\).

(b) \(\int_0^{\pi/4}\cos^4\theta\,d\theta=\frac{3\pi}{32}+\frac14\).

Let \(z=\cos\theta+i\sin\theta\), so that \(z^{-1}=\cos\theta-i\sin\theta\). Then

\(z+z^{-1}=2\cos\theta\).

Raise both sides to the fourth power:

\((z+z^{-1})^4=(2\cos\theta)^4=16\cos^4\theta\).

Expanding the left-hand side gives

\((z+z^{-1})^4=z^4+4z^2+6+4z^{-2}+z^{-4}\).

Now use de Moivre's theorem:\(z^n+z^{-n}=2\cos n\theta\). Hence

\(z^4+z^{-4}=2\cos 4\theta\) and \(z^2+z^{-2}=2\cos 2\theta\).

So

\(16\cos^4\theta=2\cos 4\theta+8\cos 2\theta+6\).

Dividing by 16 gives

\(\cos^4\theta=\frac18\cos 4\theta+\frac12\cos 2\theta+\frac38\).

Therefore

\(\int_0^{\pi/4}\cos^4\theta\,d\theta=\int_0^{\pi/4}\left(\frac18\cos 4\theta+\frac12\cos 2\theta+\frac38\right)d\theta\).

Integrating term by term:

\(=\left[\frac{\sin 4\theta}{32}+\frac{\sin 2\theta}{4}+\frac{3\theta}{8}\right]_0^{\pi/4}\).

Now \(\sin\pi=0\), \(\sin(\pi/2)=1\), so

\(=\frac14+\frac{3\pi}{32}\).