cos frac{2 pi}{7}+cos frac{4 pi}{7}+cos frac{ | vedclass

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Question

(D) Let $S = \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \pi$. Since $\cos \pi = -1$,we have $S = \cos \frac{2 \pi}{7}+\cos \f

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

(D) Let $S = \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}+\cos \pi$. Since $\cos \pi = -1$,we have $S = \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}-1$. Using the formula $\sum_{k=1}^{n} \cos(k	heta) = \frac{\sin(n	heta/2)}{\sin(	heta/2)} \cos((n+1)	heta/2)$,for the sum of cosines in arithmetic progression: $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7} = \frac{\sin(3 \pi/7)}{\sin(\pi/7)} \cos(\frac{2 \pi/7 + 6 \pi/7}{2}) = \frac{\sin(3 \pi/7)}{\sin(\pi/7)} \cos(\frac{4 \pi}{7})$. Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,the sum $C = \cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$ can be evaluated as: $2 \sin(\frac{\pi}{7}) C = 2 \sin(\frac{\pi}{7}) \cos(\frac{2 \pi}{7}) + 2 \sin(\frac{\pi}{7}) \cos(\frac{4 \pi}{7}) + 2 \sin(\frac{\pi}{7}) \cos(\frac{6 \pi}{7})$ $= (\sin \frac{3 \pi}{7} - \sin \frac{\pi}{7}) + (\sin \frac{5 \pi}{7} - \sin \frac{3 \pi}{7}) + (\sin \pi - \sin \frac{5 \pi}{7})$ $= \sin \pi - \sin \frac{\pi}{7} = 0 - \sin \frac{\pi}{7} = -\sin \frac{\pi}{7}$. Thus,$C = -\frac{1}{2}$. Therefore,$S = C - 1 = -\frac{1}{2} - 1 = -\frac{3}{2}$.

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

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