Question

The value of $\sin \frac{\pi }{7}\sin \frac{2\pi }{7}\sin \frac{3\pi }{7}$ is equal to

• A. $\frac{1}{64}$
• B. $\frac{7}{64}$
• C. $\frac{\sqrt{7}}{8}$
• D. $\frac{\sqrt{7}}{2}$

Answer 1

Answer: (C)

$\frac{\sqrt{7}}{8}$

As $\cos \frac{2\pi }{7},\cos \frac{4\pi }{7},\cos \frac{6\pi }{7}$ root of $8x^3+4x^2-4x-1=0$
Therefore
$8(x-\cos \frac{2\pi }{7})(x-\cos \frac{4\pi }{7})(x-\cos \frac{6\pi }{7})=8x^3+4x^2-4x-1$
For $x=1$
$8(1-\cos \frac{2\pi }{7})(1-\cos \frac{4\pi }{7})(1-\cos \frac{6\pi }{7})=7\Rightarrow 8(1-1+2{\sin }^2\frac{\pi }{7})(1-1+2{\sin }^2\frac{2\pi }{7})(1-1+2{\sin }^2\frac{3\pi }{7})=7\Rightarrow {\sin }^2\frac{\pi }{7}{\sin }^2\frac{2\pi }{7}{\sin }^2\frac{3\pi }{7}=\frac{7}{64}\Rightarrow \sin \frac{\pi }{7}\sin \frac{2\pi }{7}\sin \frac{3\pi }{7}=\frac{\sqrt{7}}{8}$