Q127 / 200 · WBJEE 2014 · Mathematics

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

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

Solution

Let $\frac{2 \pi r}{7}=\theta$ $\Rightarrow \quad 2 \pi r=3 \theta+4 \theta$ $\Rightarrow \quad 4 \theta=2 \pi r-3 \theta$ $\Rightarrow \quad \sin 4 \theta=\sin (2 \pi r-3 \theta)$ $\Rightarrow \quad \sin 4 \theta=-\sin 3 \theta$ $\Rightarrow 2 \sin 2 \theta \cos 2 \theta=-\left[3 \sin \theta-4 \sin ^{3} \theta\right.$ $\Rightarrow 2 \times 2 \sin \theta \cos \theta\left(2 \cos ^{2} \theta-1\right)$ $=-3 \sin \theta+4 \sin ^{3} \theta$ $\Rightarrow \sin \theta\left[8 \cos ^{3} \theta-4 \cos \theta+3-4\left(1-\cos ^{2} \theta\right)\right]=0$ $\Rightarrow 8 \cos ^{3} \theta+4 \cos ^{2} \theta-4 \cos \theta-1=0$ (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. $\therefore \quad \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}=-\frac{1}{2}$

Hence, the correct answer is option: 'C'.

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