Question

$\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$

• A. is equal to zero
• B. lies between 0 and 3
• C. is a negative number
• D. lies between 3 and 6

Answer 1

Answer: (C)

Let $\frac{2\pi r}{7}=\theta$
$\Rightarrow 2\pi r=3\theta +4\theta$
$\Rightarrow 4\theta =2\pi r-3\theta$
$\Rightarrow \sin 4\theta =\sin (2\pi r-3\theta )$
$\Rightarrow \sin 4\theta =-\sin 3\theta$
$\Rightarrow 2\sin 2\theta \cos 2\theta =-\left[3\sin \theta -4{\sin }^3\theta \right]$
$\Rightarrow 2\times 2\sin \theta \cos \theta \left(2{\cos }^2\theta -1\right)=-3\sin \theta +4{\sin }^3\theta$
$\Rightarrow \sin \theta \left[8{\cos }^3\theta -4\cos \theta +3-4\left(1-{\cos }^2\theta \right)\right]=0$
$\Rightarrow 8{\cos }^3\theta +4{\cos }^2\theta -4\cos \theta -1=0$
Thus, $\cos \frac{2\pi }{7},\cos \frac{4\pi }{7}$ and $\cos \frac{6\pi }{7}$ are the roots of the above equation.
$\therefore \cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}=-\frac{1}{2}$