Answer: (i) Let \(c=\cos\theta\) and \(s=\sin\theta\). Then \(\cos\theta+i\sin\theta=c+is\), so by de Moivre’s theorem
\((c+is)^5=\cos 5\theta+i\sin 5\theta\).
Expanding,
\((c+is)^5=c^5+5c^4(is)+10c^3(is)^2+10c^2(is)^3+5c(is)^4+(is)^5\).
Taking imaginary parts gives
\(\sin 5\theta=5c^4s-10c^2s^3+s^5\).
Since \(c^2=1-s^2\),
\(\sin 5\theta=s\big(5(1-s^2)^2-10s^2(1-s^2)+s^4\big)=5\sin\theta-20\sin^3\theta+16\sin^5\theta\).
(ii) If \(\theta=\frac{k\pi}{5}\) for \(k=0,1,2,3,4,5\), then \(\sin 5\theta=\sin k\pi=0\). So from part (i), with \(s=\sin\theta\),
\(16s^5-20s^3+5s=0\), hence \(s(16s^4-20s^2+5)=0\).
Therefore the non-zero values of \(s\) satisfy \(16s^4-20s^2+5=0\). Taking \(\theta=\frac{\pi}{5}\) and \(\theta=\frac{2\pi}{5}\) gives roots \(s=\pm\sin\frac{\pi}{5}\) and \(s=\pm\sin\frac{2\pi}{5}\).
(iii) Using \(\sin\frac{3\pi}{5}=\sin\frac{2\pi}{5}\) and \(\sin\frac{4\pi}{5}=\sin\frac{\pi}{5}\),
\(\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5}\).
Now the numbers \(\sin^2\frac{\pi}{5}\) and \(\sin^2\frac{2\pi}{5}\) are the two positive roots of \(16t^2-20t+5=0\) in \(t=s^2\). Hence their product is the constant term divided by the leading coefficient:
\(\sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5}=\frac{5}{16}\).
So \(\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\frac{5}{16}\).
Also, for the quadratic \(16t^2-20t+5=0\), the sum of the roots is \(\frac{20}{16}=\frac54\). Therefore
\(\sin^2\frac{\pi}{5}+\sin^2\frac{2\pi}{5}=\frac54\).
(i) Write \(\cos\theta=c\) and \(\sin\theta=s\). Then
\(c+is=\cos\theta+i\sin\theta\), so by de Moivre’s theorem
\((c+is)^5=\cos 5\theta+i\sin 5\theta\).
Expanding by the binomial theorem gives
\(c^5+5c^4(is)+10c^3(is)^2+10c^2(is)^3+5c(is)^4+(is)^5\).
Taking imaginary parts,
\(\sin 5\theta=5c^4s-10c^2s^3+s^5\).
Now substitute \(c^2=1-s^2\):
\(\sin 5\theta=s\big(5(1-s^2)^2-10s^2(1-s^2)+s^4\big)\).
Expand the bracket:
\(5(1-2s^2+s^4)-10s^2+10s^4+s^4=5-20s^2+16s^4\).
Hence
\(\sin 5\theta=s(5-20s^2+16s^4)=5\sin\theta-20\sin^3\theta+16\sin^5\theta\).
(ii) Let \(s=\sin\theta\). If \(\theta=0,\pm\frac{\pi}{5},\pm\frac{2\pi}{5}\), then \(5\theta\) is a multiple of \(\pi\), so \(\sin 5\theta=0\). Using part (i),
\(16s^5-20s^3+5s=0\).
Factor out \(s\):
\(s(16s^4-20s^2+5)=0\).
So either \(s=0\) or \(16s^4-20s^2+5=0\). The non-zero values of \(s\) occur when \(\theta=\pm\frac{\pi}{5},\pm\frac{2\pi}{5}\), giving the roots
\(x=\pm\sin\frac{\pi}{5},\ \pm\sin\frac{2\pi}{5}\).
(iii) Since \(\sin\frac{3\pi}{5}=\sin\frac{2\pi}{5}\) and \(\sin\frac{4\pi}{5}=\sin\frac{\pi}{5}\),
\(\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5}\).
From part (ii), the values \(t=\sin^2\frac{\pi}{5}\) and \(t=\sin^2\frac{2\pi}{5}\) satisfy
\(16t^2-20t+5=0\).
For this quadratic, the product of the roots is \(\frac{5}{16}\). Therefore
\(\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5}=\frac{5}{16}\).
The sum of the roots is \(\frac{20}{16}=\frac54\), so
\(\sin^2\frac{\pi}{5}+\sin^2\frac{2\pi}{5}=\frac54\).
