Answer: (a) \(\displaystyle \operatorname{cosec} 5\theta=\frac{\operatorname{cosec}^{5}\theta}{5\operatorname{cosec}^{4}\theta-20\operatorname{cosec}^{2}\theta+16}\).
(b) The roots are
\(\displaystyle x=\operatorname{cosec}\left(\frac{\pi}{30}\right),\ \operatorname{cosec}\left(\frac{5\pi}{30}\right),\ \operatorname{cosec}\left(-\frac{7\pi}{30}\right),\ \operatorname{cosec}\left(-\frac{11\pi}{30}\right),\ \operatorname{cosec}\left(\frac{13\pi}{30}\right).\)
(a) Let \(c=\cos\theta\) and \(s=\sin\theta\). By de Moivre's theorem,
\(\cos 5\theta+i\sin 5\theta=(c+is)^5.\)
Expanding and taking the imaginary part,
\(\sin 5\theta=s^5-10c^2s^3+5c^4s.\)
Using \(c^2=1-s^2\),
\(\sin 5\theta=s^5-10(1-s^2)s^3+5(1-s^2)^2s.\)
So
\(\sin 5\theta=s^5-10s^3+10s^5+5s-10s^3+5s^5=16s^5-20s^3+5s.\)
Hence
\(\operatorname{cosec}5\theta=\frac{1}{16s^5-20s^3+5s}.\)
Now multiply top and bottom by \(\operatorname{cosec}^5\theta=\dfrac{1}{s^5}\):
\(\operatorname{cosec}5\theta=\frac{\operatorname{cosec}^5\theta}{16-20\operatorname{cosec}^2\theta+5\operatorname{cosec}^4\theta}\)
which may be written as
\(\operatorname{cosec} 5\theta=\frac{\operatorname{cosec}^{5} \theta}{5 \operatorname{cosec}^{4} \theta-20 \operatorname{cosec}^{2} \theta+16}.\)
(b) Rearranging
\(x^5-10x^4+40x^2-32=0\)
gives
\(x^5=2(5x^4-20x^2+16),\)
so
\(\frac{x^5}{5x^4-20x^2+16}=2.\)
If \(x=\operatorname{cosec}\theta\), then from part (a)
\(\frac{x^5}{5x^4-20x^2+16}=\operatorname{cosec}5\theta.\)
Therefore
\(\operatorname{cosec}5\theta=2,\)
so
\(\sin 5\theta=\frac12.\)
Thus
\(5\theta=\frac{\pi}{6}+2k\pi\quad\text{or}\quad 5\theta=\frac{5\pi}{6}+2k\pi.\)
Hence
\(\theta=\frac{\pi}{30}+\frac{2k\pi}{5}\quad\text{or}\quad \theta=\frac{\pi}{6}+\frac{2k\pi}{5}.\)
Taking five distinct values that give the five roots, one suitable set is
\(\theta=\frac{\pi}{30},\ \frac{5\pi}{30},\ -\frac{7\pi}{30},\ -\frac{11\pi}{30},\ \frac{13\pi}{30}.\)
Therefore the roots are
\(\displaystyle x=\operatorname{cosec}\left(\frac{\pi}{30}\right),\ \operatorname{cosec}\left(\frac{5\pi}{30}\right),\ \operatorname{cosec}\left(-\frac{7\pi}{30}\right),\ \operatorname{cosec}\left(-\frac{11\pi}{30}\right),\ \operatorname{cosec}\left(\frac{13\pi}{30}\right).\)
