Question

Prove that :
$\sin \frac{\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\sin \frac{3\mathrm{\pi }}{5}\sin \frac{4\mathrm{\pi }}{5}=\frac{5}{16}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\sin \frac{\mathrm{\pi }}{5}\sin \frac{2\mathrm{\pi }}{5}\sin \frac{3\mathrm{\pi }}{5}\sin \frac{4\pi }{5} \\ =\frac{1}{2}\left(2\sin \frac{\mathrm{\pi }}{5}\sin \frac{4\mathrm{\pi }}{5}\right)\frac{1}{2}\left(2\sin \frac{2\mathrm{\pi }}{5}\sin \frac{3\mathrm{\pi }}{5}\right) \\ =\frac{1}{4}\left(\cos \left(\frac{\mathrm{\pi }}{5}-\frac{4\mathrm{\pi }}{5}\right)-\cos \left(\frac{\mathrm{\pi }}{5}+\frac{4\mathrm{\pi }}{5}\right)\right)\left(\cos \left(\frac{2\mathrm{\pi }}{5}-\frac{3\mathrm{\pi }}{5}\right)-\cos \left(\frac{2\mathrm{\pi }}{5}+\frac{3\mathrm{\pi }}{5}\right)\right) \\ =\frac{1}{4}\left(\cos \left(\frac{-3\mathrm{\pi }}{5}\right)-\cos \left(\frac{5\mathrm{\pi }}{5}\right)\right)\left(\cos \left(\frac{-\mathrm{\pi }}{5}\right)-\cos \left(\frac{5\mathrm{\pi }}{5}\right)\right) \\ =\frac{1}{4}\left(\cos \left(\frac{3\mathrm{\pi }}{5}\right)-\cos \left(\mathrm{\pi }\right)\right)\left(\cos \left(\frac{\mathrm{\pi }}{5}\right)-\cos \left(\mathrm{\pi }\right)\right) \\ =\frac{1}{4}\left(\cos \left(\frac{3\mathrm{\pi }}{5}\right)+1\right)\left(\cos \left(\frac{\mathrm{\pi }}{5}\right)+1\right) \\ =\frac{1}{4}\left(\cos \left(\mathrm{\pi }-\frac{2\mathrm{\pi }}{5}\right)+1\right)\left(\cos \left(\frac{\mathrm{\pi }}{5}\right)+1\right) \\ =\frac{1}{4}\left(-\cos \left(\frac{2\mathrm{\pi }}{5}\right)+1\right)\left(\left(\frac{\sqrt{5}+1}{4}\right)+1\right)\left(\because \cos \frac{\mathrm{\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)\left(\because \cos \frac{2\mathrm{\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{{\left(\sqrt{5}\right)}^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} \\ =\mathrm{RHS} \end{gathered}$$

Thus, LHS = RHS

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