# The value of sin

Question:

The value of $\sin \frac{\pi}{10} \sin \frac{13 \pi}{10}$ is

(a) $\frac{1}{2}$

(b) $-\frac{1}{2}$

(c) $-\frac{1}{4}$

(d) 1

Solution:

$\sin \frac{\pi}{10} \sin \frac{13 \pi}{10}$

$=\sin \frac{\pi}{10} \sin \left(\pi+\frac{3 \pi}{10}\right)$

$=\sin \frac{\pi}{10}\left(-\sin \frac{3 \pi}{10}\right) \quad(\because \sin (\pi+\theta)=-\sin \theta)$

$\therefore \sin \frac{\pi}{10} \sin \frac{13 \pi}{10}=-\sin \frac{\pi}{10} \sin \frac{3 \pi}{10} \quad \ldots$ (1)

Let $\theta=\frac{\pi}{10} \quad$ i. e. $\quad 5 \theta=\frac{\pi}{2}$

i.e. $3 \theta+2 \theta=\frac{\pi}{2}$

i. e. $3 \theta=\frac{\pi}{2}-2 \theta$

i. e. $\cos 3 \theta=\cos \left(\frac{\pi}{2}-2 \theta\right)=\sin 2 \theta$

i. e. $4 \cos ^{3} \theta-3 \cos \theta=2 \sin \theta \cos \theta$(using identity of $\cos 3 \theta$ and $\sin 2 \theta$ )

i.e. $4 \cos ^{2} \theta-3=2 \sin \theta$

i.e. $4\left(1-\sin ^{2} \theta\right)-3=2 \sin \theta$

i. e. $4 \sin ^{2} \theta+2 \sin \theta-1=0 \quad\left(\right.$ i. e. Let $\sin \theta=t 4 t^{2}+2 t-1$ roots are $\left.=\frac{-2 \pm \sqrt{20}}{2 \times 4}=\pm\left(\frac{\sqrt{5}-1}{4}\right)\right)$

i. e. $\sin \theta=\frac{\sqrt{5}-1}{4}$

i. e. $\sin \frac{\pi}{10}=\frac{\sqrt{5}-1}{4}$

$\sin \frac{3 \pi}{10}=3 \sin \frac{\pi}{10}-4 \sin ^{3} \frac{\pi}{10}$

$=3\left(\frac{\sqrt{5}-1}{4}\right)-4\left(\frac{\sqrt{5}-1}{4}\right)^{3}$

$=\frac{3}{4}(\sqrt{5}-1)-\frac{1}{4^{2}}(\sqrt{5}-1)^{3}$

$=\frac{3}{4}(\sqrt{5}-1)-\frac{1}{16}\left[(\sqrt{5})^{3}-(1)^{3}-3 \sqrt{5}(\sqrt{5}-1)\right]$

$=\frac{3}{4} \sqrt{5}-\frac{3}{4}-\frac{1}{16}[5 \sqrt{5}-1-3 \times 5+3 \sqrt{5}]$

$=\frac{3}{4} \sqrt{5}-\frac{3}{4}-\frac{1}{16}[5 \sqrt{5}-1-15+3 \sqrt{5}]$

$=\frac{3}{4} \sqrt{5}-\frac{3}{4}-\frac{1}{16}[8 \sqrt{5}-16]$

$=\frac{3}{4} \sqrt{5}-\frac{3}{4}-\frac{1}{2} \sqrt{5}+1$

$=\frac{3}{4} \sqrt{5}-\frac{1}{2} \sqrt{5}+1-\frac{3}{4}$

$=\frac{3 \sqrt{5}-2 \sqrt{5}}{4}+\frac{4-3}{4}$

$=\frac{\sqrt{5}}{4}+\frac{1}{4}$

$\sin \frac{3 \pi}{10}=\frac{1}{4}(\sqrt{5}+1)$

from (1)

$\therefore \sin \frac{\pi}{10} \sin \frac{13 \pi}{10}=-\sin \frac{\pi}{10} \sin \frac{3 \pi}{10}$

$=-\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)$

$=-\left(\frac{5-1}{4^{2}}\right)=\frac{-4}{16}=\frac{-1}{4}$

Hence $\sin \frac{13 \pi}{10} \sin \frac{\pi}{10}=\frac{-1}{4}$