Question

Prove that:
$\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{7\pi }{15}=\frac{1}{16}$

Answer 1

$\mathrm{LHS}=\cos \frac{\mathrm{\pi }}{15}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{7\mathrm{\pi }}{15}$

$$\begin{gathered} =\frac{2\sin {\displaystyle \frac{\mathrm{\pi }}{15}}\cos {\displaystyle \frac{\mathrm{\pi }}{15}}}{2\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\cos \frac{2\mathrm{\pi }}{15}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{7\mathrm{\pi }}{15} \\ \left[\mathrm{On}\mathrm{dividing}\mathrm{and}\mathrm{multiplying}\mathrm{by}2\sin \frac{\mathrm{\pi }}{15}\right] \end{gathered}$$

$$\begin{gathered} =\frac{{\displaystyle 2\sin \frac{2\mathrm{\pi }}{15}\times \cos \frac{2\mathrm{\pi }}{15}}}{2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\cos \frac{4\mathrm{\pi }}{15}\cos \frac{7\mathrm{\pi }}{15} \\ =\frac{{\displaystyle 2\sin \frac{4\mathrm{\pi }}{15}\times \cos \frac{4\mathrm{\pi }}{15}}}{2\times 2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\cos \frac{7\mathrm{\pi }}{15} \\ =\frac{{\displaystyle \sin \frac{8\mathrm{\pi }}{15}}}{2\times 2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\cos \frac{7\mathrm{\pi }}{15} \end{gathered}$$

$$\begin{gathered} =\frac{{\displaystyle 2\sin \frac{8\mathrm{\pi }}{15}\cos \frac{7\mathrm{\pi }}{15}}}{2\times 2\times 2\times 2\sin {\displaystyle \frac{\mathrm{\pi }}{15}}} \\ =\frac{{\displaystyle 2\sin \frac{8\mathrm{\pi }}{15}\cos \frac{7\mathrm{\pi }}{15}}}{16\sin {\displaystyle \frac{\mathrm{\pi }}{15}}} \\ =\frac{{\displaystyle \sin \left(\frac{8\mathrm{\pi }}{15}+\frac{7\mathrm{\pi }}{15}\right)+}\sin \left({\displaystyle \frac{8\mathrm{\pi }}{15}}-{\displaystyle \frac{7\mathrm{\pi }}{15}}\right)}{16\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}\left[\because 2\sin A\cos B=\sin \left(A+B\right)+\sin \left(A-B\right)\right] \end{gathered}$$

$$\begin{gathered} =\frac{{\displaystyle \mathrm{sin\pi }+}\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}{16\sin {\displaystyle \frac{\mathrm{\pi }}{15}}} \\ =\frac{0{\displaystyle +}\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}{16\sin {\displaystyle \frac{\mathrm{\pi }}{15}}} \\ =\frac{{\displaystyle }\sin {\displaystyle \frac{\mathrm{\pi }}{15}}}{16\sin {\displaystyle \frac{\mathrm{\pi }}{15}}} \\ =\frac{1}{16} \\ =\mathrm{RHS} \end{gathered}$$
Hence proved.