Question

Prove that $\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{8\pi }{7}=\frac{1}{8}$.

Answer 1

Multiply both sides with $sin\left(\frac{\pi }{7}\right)$ hence we have that

$sin\left(\frac{\pi }{7}\right)cos\left(\frac{2\pi }{7}\right)+sin\left(\frac{\pi }{7}\right)cos\left(\frac{4\pi }{7}\right)+sin\left(\frac{\pi }{7}\right)\cdot cos\left(\frac{6\pi }{7}\right)=-\frac{1}{2}\ast sin\left(\frac{\pi }{7}\right)$

Then because $sinA\cdot cosB=\frac{1}{2}\left[sin\right(A-B)+sin(A+B\left)\right]$ were write this as

$\frac{1}{2}[sin(\frac{\pi }{7}-\frac{2\pi }{7})+sin(\frac{\pi }{7}+\frac{2\pi }{7})]+\frac{1}{2}[sin(\frac{\pi }{7}-\frac{4\pi }{7})+sin(\frac{\pi }{7}+\frac{4\pi }{7})]+\frac{1}{2}[sin(\frac{\pi }{7}-\frac{6\pi }{7})+sin(\frac{\pi }{7}+\frac{6\pi }{7})]=\frac{-sin\frac{\pi }{7}}{2}$

We'l ldivide by $\frac{1}{2}$ both sides:

$-sin\left(\frac{\pi }{7}\right)+sin\left(\frac{3\pi }{7}\right)-sin\left(\frac{3\pi }{7}\right)+sin\left(\frac{5\pi }{7}\right)-sin\left(\frac{5\pi }{7}\right)+sin\left(\frac{7\pi }{7}\right)=-sin\left(\frac{\pi }{7}\right)$

Cancelling the same terms yields to $-sin\left(\frac{\pi }{7}\right)+sin\left(\pi \right)=-sin\left(\frac{\pi }{7}\right)$

But $sin\left(\pi \right)=0$hence

$-sin\left(\frac{\pi }{7}\right)=-sin\left(\frac{\pi }{7}\right)$

Since we have the same results in both sides,the given identity is true.