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)

Is a negative number

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