Question

Let $\alpha =\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$ and $\beta ={\sin }^2\frac{\pi }{8}+{\sin }^2\frac{3\pi }{8}+{\sin }^2\frac{5\pi }{8}+{\sin }^2\frac{7\pi }{8}$. Then the value of $\alpha \beta$ is

• A. $2$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $1$

Answer 1

The correct option is B $-1$

$\alpha =\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$
$=\cos \theta +\cos 2\theta +\cos 3\theta ,$ where $\theta =\frac{2\pi }{7}$
$=2\cos 2\theta \cos \theta +\cos 2\theta$
$=\frac{\sin 2\theta \cos 2\theta }{\sin \theta }+\cos 2\theta$
$=\frac{\sin 4\theta +2\cos 2\theta \sin \theta }{2\sin \theta }$
$=\frac{\sin 4\theta +\sin 3\theta -\sin \theta }{2\sin \theta }$
$\Rightarrow \alpha =-\frac{1}{2}$
$(\because \sin 4\theta +\sin 3\theta =\sin 4\theta -\sin 4\theta =0$ as $7\theta =2\pi )$

$\beta ={\sin }^2\frac{\pi }{8}+{\sin }^2\frac{3\pi }{8}+{\sin }^2\frac{5\pi }{8}+{\sin }^2\frac{7\pi }{8}$ $=2({\sin }^2\frac{\pi }{8}+{\sin }^2\frac{3\pi }{8})\left\{\begin{array}{c}\because \sin \frac{7\pi }{8}=\sin (\pi -\frac{\pi }{8})=\sin \frac{\pi }{8}\\ \sin \frac{5\pi }{8}=\sin (\pi -\frac{3\pi }{8})=\sin \frac{3\pi }{8}\end{array}\right\}$
$=2({\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8})\{\because \sin \frac{3\pi }{8}=\sin (\frac{\pi }{2}-\frac{\pi }{8})=\cos \frac{\pi }{8}\}$
$\Rightarrow \beta =2$

Now, $\alpha \beta =-1$