# Prove that:

Question:

Prove that:

$\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5} \sin \frac{4 \pi}{5}=\frac{5}{16}$

Solution:

LHS $=\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5} \sin \frac{4 \pi}{5}$

$=\frac{1}{2}\left(2 \sin \frac{\pi}{5} \sin \frac{4 \pi}{5}\right) \frac{1}{2}\left(2 \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5}\right)$

$=\frac{1}{4}\left(\cos \left(\frac{\pi}{5}-\frac{4 \pi}{5}\right)-\cos \left(\frac{\pi}{5}+\frac{4 \pi}{5}\right)\right)\left(\cos \left(\frac{2 \pi}{5}-\frac{3 \pi}{5}\right)-\cos \left(\frac{2 \pi}{5}+\frac{3 \pi}{5}\right)\right)$

$=\frac{1}{4}\left(\cos \left(\frac{-3 \pi}{5}\right)-\cos \left(\frac{5 \pi}{5}\right)\right)\left(\cos \left(\frac{-\pi}{5}\right)-\cos \left(\frac{5 \pi}{5}\right)\right)$

$=\frac{1}{4}\left(\cos \left(\frac{3 \pi}{5}\right)-\cos (\pi)\right)\left(\cos \left(\frac{\pi}{5}\right)-\cos (\pi)\right)$

$=\frac{1}{4}\left(\cos \left(\frac{3 \pi}{5}\right)+1\right)\left(\cos \left(\frac{\pi}{5}\right)+1\right)$

$=\frac{1}{4}\left(\cos \left(\pi-\frac{2 \pi}{5}\right)+1\right)\left(\cos \left(\frac{\pi}{5}\right)+1\right)$

$=\frac{1}{4}\left(-\cos \left(\frac{2 \pi}{5}\right)+1\right)\left(\left(\frac{\sqrt{5}+1}{4}\right)+1\right) \quad\left(\because \cos \frac{\pi}{5}=\frac{\sqrt{5}+1}{4}\right)$

$=\frac{1}{4}\left(-\left(\frac{\sqrt{5}-1}{4}\right)+1\right)\left(\left(\frac{\sqrt{5}+1}{4}\right)+1\right) \quad\left(\because \cos \frac{2 \pi}{5}=\frac{\sqrt{5}-1}{4}\right)$

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

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

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

$=\frac{1}{4}\left(-\frac{1}{4}+\frac{2}{4}+1\right)$

$=\frac{1}{4}\left(\frac{-1+2+4}{4}\right)$

$=\frac{5}{16}$

= RHS

Thus, LHS = RHS

Hence, $\sin \frac{\pi}{5} \sin \frac{2 \pi}{5} \sin \frac{3 \pi}{5} \sin \frac{4 \pi}{5}=\frac{5}{16}$.

Leave a comment

Click here to get exam-ready with eSaral