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.
$\therefore \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=-\frac{1}{2}$