Answer: a) \(\cos^4\theta=\frac18\bigl(\cos 4\theta+4\cos 2\theta+3\bigr)\)

b) \(\sin^4\theta=\frac18\bigl(\cos 4\theta-4\cos 2\theta+3\bigr)\)

Hence \(\displaystyle \int_0^{\pi/8}(\cos^4\theta+\sin^4\theta)\,d\theta=\frac{1}{16}+\frac{3\pi}{32}\).

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

Then

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

so

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

Now \(z+z^{-1}=2\cos\theta\), \(z^2+z^{-2}=2\cos 2\theta\), and \(z^4+z^{-4}=2\cos 4\theta\). Hence

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

Dividing by \(16\) gives

\(\cos^4\theta=\frac{1}{16}(2\cos 4\theta+8\cos 2\theta+6)=\frac18\bigl(\cos 4\theta+4\cos 2\theta+3\bigr).\)

For \(\sin^4\theta\), use \(z-z^{-1}=2i\sin\theta\). Expanding similarly,

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

so

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

Therefore

\((2i\sin\theta)^4=2\cos 4\theta-8\cos 2\theta+6.\)

Since \((2i)^4=16\), this becomes

\(16\sin^4\theta=2\cos 4\theta-8\cos 2\theta+6,\)

hence

\(\sin^4\theta=\frac{1}{16}(2\cos 4\theta-8\cos 2\theta+6)=\frac18\bigl(\cos 4\theta-4\cos 2\theta+3\bigr).\)

Now add the two expressions:

\(\cos^4\theta+\sin^4\theta=\frac18\bigl(2\cos 4\theta+6\bigr)=\frac14(\cos 4\theta+3).\)

So

\(\displaystyle \int_0^{\pi/8}(\cos^4\theta+\sin^4\theta)\,d\theta=\frac14\int_0^{\pi/8}(\cos 4\theta+3)\,d\theta.\)

Integrating,

\(\frac14\left[\frac{\sin 4\theta}{4}+3\theta\right]_0^{\pi/8}.\)

At \(\theta=\pi/8\), \(\sin(\pi/2)=1\), so

\(\frac14\left(\frac14+\frac{3\pi}{8}\right)=\frac{1}{16}+\frac{3\pi}{32}.\)