Question

The value of $\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{3\pi }{15}\cos \frac{4\pi }{15}\cos \frac{5\pi }{15}\cos \frac{6\pi }{15}\cos \frac{7\pi }{15}$ is equal to

• A. $\frac{1}{2^6}$
• B. $\frac{1}{2^8}$
• C. $\frac{1}{2^7}$
• D. None of these

Answer 1

The correct option is A $\frac{1}{2^6}$

Solving using continue product,

We know that,

$cosx.cos2x........cosnx=\frac{sin{2}^{n}x}{2^n.sinx}$

Using this, we get

$⟹\frac{sin{2}^7\left(\frac{\pi }{15}\right)}{2^{7}sin\left(\frac{\pi }{15}\right)}$

$⟹\frac{2}{2^7}$

$⟹\frac{1}{2^6}$