Question

The expression $\cos A\cos 2A\cos 2^{2}A....\cos 2^{n-1}A$ $=\{\begin{array}{cc}l\frac{\sin 2^{n}A}{2^n\sin A}& \text{if}A\ne n\pi \\ \frac{1}{2^n}& \text{if}A=\frac{\pi }{2^n+1}\forall n\in Z\\ -\frac{1}{2^n}& \text{if}A=\frac{\pi }{2^n-1}\end{array}$

The value of $\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\cos \frac{14\pi }{15}$ is equal to

• A. $\frac{-1}{16}$
• B. $\frac{1}{16}$
• C. $\frac{1}{32}$
• D. $\frac{1}{8}$

Answer 1

Answer: (B)

$\frac{1}{16}$

We have, $cos\frac{2\pi }{15}cos\frac{4\pi }{15}cos\frac{8\pi }{15}cos\frac{14\pi }{15}$

$=cos\frac{2\pi }{15}cos\frac{4\pi }{15}cos\frac{8\pi }{15}cos(\pi -\frac{\pi }{15})$

$=-cos\frac{\pi }{15}cos\frac{2\pi }{15}cos\frac{4\pi }{15}cos\frac{8\pi }{15}$

Using formula given in passage for $n=4$

$=-\frac{sin{2}^4\frac{\pi }{15}}{2^{4}sin\frac{\pi }{15}}=-\frac{sin\frac{16\pi }{15}}{16sin\frac{\pi }{15}}=-\frac{sin(\pi +\frac{\pi }{15})}{16sin\frac{\pi }{15}}$

$=\frac{sin\frac{\pi }{15}}{16sin\frac{\pi }{15}}=\frac{1}{16}$