(a) Let \(z=\cos\theta+\mathrm{i}\sin\theta\). Then \(z^{-1}=\cos\theta-\mathrm{i}\sin\theta\), so
\(z+z^{-1}=2\cos\theta\).
Now expand:
\((z+z^{-1})^4=z^4+4z^2+6+4z^{-2}+z^{-4}\)
and group the terms as
\((z+z^{-1})^4=(z^4+z^{-4})+4(z^2+z^{-2})+6\).
By de Moivre's theorem, \(z^n=\cos n\theta+\mathrm{i}\sin n\theta\), hence
\(z^n+z^{-n}=2\cos n\theta\).
So
\(z^4+z^{-4}=2\cos4\theta,\qquad z^2+z^{-2}=2\cos2\theta.\)
Also \((z+z^{-1})^4=(2\cos\theta)^4=16\cos^4\theta\). Therefore
\(16\cos^4\theta=2\cos4\theta+4(2\cos2\theta)+6\)
\(\phantom{16\cos^4\theta}=2\cos4\theta+8\cos2\theta+6.\)
Dividing by \(16\),
\(\cos^4\theta=\frac{1}{8}(\cos4\theta+4\cos2\theta+3).\)
(b) Use \(x=\sin\theta\). Then
\(dx=\cos\theta\,d\theta\), and when \(x=0\), \(\theta=0\); when \(x=\frac12\), \(\theta=\frac\pi6\).
Also,
\(1-x^2=1-\sin^2\theta=\cos^2\theta,\)
so on \([0,\pi/6]\),
\((1-x^2)^{3/2}=(\cos^2\theta)^{3/2}=\cos^3\theta.\)
Hence
\(\displaystyle \int_0^{1/2}(1-x^2)^{3/2}\,dx=\int_0^{\pi/6}\cos^3\theta\cdot\cos\theta\,d\theta=\int_0^{\pi/6}\cos^4\theta\,d\theta.\)
Using part (a),
\(\cos^4\theta=\frac18(\cos4\theta+4\cos2\theta+3),\)
so
\(\displaystyle \int_0^{\pi/6}\cos^4\theta\,d\theta=\frac18\int_0^{\pi/6}(\cos4\theta+4\cos2\theta+3)\,d\theta\)
\(\displaystyle =\frac18\left[\frac14\sin4\theta+2\sin2\theta+3\theta\right]_0^{\pi/6}.\)
Now \(\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}\) and \(\sin\frac\pi3=\frac{\sqrt3}{2}\), so
\(\displaystyle \frac18\left(\frac14\cdot\frac{\sqrt3}{2}+2\cdot\frac{\sqrt3}{2}+\frac{3\pi}{6}\right)\)
\(\displaystyle =\frac18\left(\frac{\sqrt3}{8}+\sqrt3+\frac\pi2\right)\)
\(\displaystyle =\frac18\left(\frac{9\sqrt3}{8}+\frac\pi2\right)\)
\(\displaystyle =\frac{9\sqrt3}{64}+\frac\pi{16}=\frac{1}{16}\left(\frac{9\sqrt3}{4}+\pi\right).\)
