Question

If $\alpha \frac{\pi }{15}$,then value of $\prod _{r-1}^7\cos r\alpha$ is equal to

• A. $\frac{-1}{64}$
• B. $\frac{1}{64}$
• C. $\frac{1}{128}$
• D. $\frac{-1}{128}$

Answer 1

The correct option is C $\frac{1}{128}$

${\prod }_{r=1}^7\cos r\alpha =\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{3\pi }{15}\cos \frac{4\pi }{15}\cos \frac{5\pi }{15}\cos \frac{6\pi }{15}\cos \frac{7\pi }{15}=\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{\pi }{5}\cos \frac{4\pi }{15}\cos \frac{\pi }{3}\cos \frac{2\pi }{5}\cos (\pi -\frac{8\pi }{15})=-\frac{1}{2}\left(\cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}\right)\left(\cos \frac{\pi }{5}\cos \frac{2\pi }{5}\right)$
As $\frac{\pi }{15}=\frac{1}{2^4-1}\Rightarrow \cos \frac{\pi }{15}\cos \frac{2\pi }{15}\cos \frac{4\pi }{15}\cos \frac{8\pi }{15}=-\frac{1}{2^4}=-\frac{1}{16}$
and $\frac{\pi }{5}=\frac{1}{2^2+1}\Rightarrow \cos \frac{\pi }{5}\cos \frac{2\pi }{5}=\frac{1}{2^2}=\frac{1}{4}$
Therefore,
${\prod }_{r=1}^7\cos r\alpha =-\frac{1}{2}(-\frac{1}{16})\left(\frac{1}{4}\right)=\frac{1}{128}$