Question

If $\alpha =\frac{\pi }{13}$, then the value of ${\Pi }_{r=1}^6\cos r\alpha$ is:

• A. $\frac{1}{64}$
• B. $-\frac{1}{64}$
• C. $\frac{1}{32}$
• D. $-\frac{1}{8}$

Answer 1

The correct option is A $\frac{1}{64}$

Let $A={\Pi }_{r=1}^6\cos \left(r\alpha \right)$. Then
$A=\cos \alpha .\cos 2\alpha .\cos 3\alpha .\cos 4\alpha .\cos 5\alpha .\cos 6\alpha .....\left(i\right)$
Where $\alpha =\frac{\pi }{13}$, then $13\alpha =\pi$, and $\alpha =\pi -12\alpha ,5\alpha =\pi -8\alpha$
$\Rightarrow \cos \alpha =-\cos 12\alpha ,\cos \left(5\alpha \right)=-\cos 8\alpha$
Now (1) $\Rightarrow A=\cos 12\alpha .\cos 2\alpha .\cos 3\alpha .\cos 4\alpha .\cos 8\alpha .\cos 6\alpha$
$\Rightarrow A=(\cos 2\alpha .\cos 4\alpha .\cos 8\alpha )(\cos 3\alpha .\cos 6\alpha .\cos 12\alpha )$
$=(\cos \theta .\cos 2\theta .\cos 4\theta )(\cos \varphi .\cos 2\varphi .\cos 4\varphi )$, where $\theta =2\alpha ,\varphi =3\alpha$
$=\frac{\sin \left(2^3\theta \right)}{2^3\sin \theta )}.\frac{\sin \left(2^3\varphi \right)}{2^3\sin \varphi }=\frac{1}{64}.\frac{\sin \left(16\alpha \right).\sin \left(24\alpha \right)}{\sin \left(2\alpha \right).\sin \left(3/alpha\right)}....\left(ii\right)$
But $\sin \left(6\alpha \right)=\sin (\pi +3\alpha +=-\sin 3\alpha ,\sin (24\alpha )=\sin (2\pi -2\alpha )=-2\alpha$
$\Rightarrow \sin \left(16\alpha \right)\sin \left(24\alpha \right)=\left(\sin 2\alpha \right).\left(\sin 3\alpha \right)$ putting in (ii), we get $A=\frac{1}{64}$.
Option A