Question

Find the value of -$\cos \frac{\pi }{11}\cos \frac{2\pi }{11}\cos \frac{3\pi }{11}\cos \frac{4\pi }{11}\cos \frac{5\pi }{11}$.

• A. -$\frac{1}{4}$
• B. -$\frac{1}{8}$
• C. -$\frac{1}{16}$
• D. -$\frac{1}{32}$

Answer 1

The correct option is D -$\frac{1}{32}$

$\cos (\pi -\frac{3\pi }{11})=-\cos \frac{8\pi }{11}$
$\cos (\pi +\frac{5\pi }{11})=-\cos \frac{16\pi }{11}$
The given expression would become
$\cos \frac{\pi }{11}\cos \frac{2\pi }{11}\cos \frac{4\pi }{11}\cos \frac{8\pi }{11}\cos \frac{16\pi }{11}$
Let $A=\frac{\pi }{11}$. Then the equation becomes
$=\cos A\cos 2A\cos 2^{2}A\cos 2^{3}A\cos 2^{4}A$
Using the formula
$cosAcos2Acos\left(2^{2}A\right)cos\left(2^{3}A\right)........cos\left(2^{n-1}A\right)=\frac{sin{2}^{n}A}{2^{n}sinA}$
Thus,
$=\frac{sin{2}^{5}A}{2^{5}sinA}$
$=\frac{\sin 32A}{32\sin A}=\frac{\sin (33A-A)}{32\sin A}$............................$\because 11A=\pi$
$=\frac{\sin (3\pi -A)}{32\sin A}$
$=\left(\frac{\sin A}{32\sin A}\right)$
Hence, option 'D' is correct.
$=\frac{1}{32}$