The value of cos left(frac{2 pi}{7}right) + c | vedclass

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Question

(A) Let $S = \cos \left(\frac{2 \pi}{7}\right) + \cos \left(\frac{4 \pi}{7}\right) + \cos \left(\frac{6 \pi}{7}\right)$.Using the formula for the sum

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Let $S = \cos \left(\frac{2 \pi}{7}ight) + \cos \left(\frac{4 \pi}{7}ight) + \cos \left(\frac{6 \pi}{7}ight)$.Using the formula for the sum of cosines in arithmetic progression: $\sum_{k=1}^{n} \cos(a + (k-1)d) = \frac{\sin(nd/2)}{\sin(d/2)} \cos\left(a + \frac{(n-1)d}{2}ight)$.Here,$a = \frac{2 \pi}{7}$,$d = \frac{2 \pi}{7}$,and $n = 3$.$S = \frac{\sin(3 	imes \frac{2 \pi}{7} / 2)}{\sin(\frac{2 \pi}{7} / 2)} \cos\left(\frac{2 \pi}{7} + \frac{(3-1) 	imes \frac{2 \pi}{7}}{2}ight)$$S = \frac{\sin(\frac{3 \pi}{7})}{\sin(\frac{\pi}{7})} \cos\left(\frac{2 \pi}{7} + \frac{2 \pi}{7}ight)$$S = \frac{\sin(\frac{3 \pi}{7})}{\sin(\frac{\pi}{7})} \cos\left(\frac{4 \pi}{7}ight)$Multiply numerator and denominator by $2 \sin(\frac{\pi}{7})$:$S = \frac{2 \sin(\frac{3 \pi}{7}) \cos(\frac{4 \pi}{7})}{2 \sin(\frac{\pi}{7})}$Using $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$S = \frac{\sin(\frac{3 \pi}{7} + \frac{4 \pi}{7}) + \sin(\frac{3 \pi}{7} - \frac{4 \pi}{7})}{2 \sin(\frac{\pi}{7})}$$S = \frac{\sin(\pi) + \sin(-\frac{\pi}{7})}{2 \sin(\frac{\pi}{7})}$Since $\sin(\pi) = 0$ and $\sin(-	heta) = -\sin(	heta)$:$S = \frac{0 - \sin(\frac{\pi}{7})}{2 \sin(\frac{\pi}{7})} = -\frac{1}{2}$.

The value of $\cos \left(\frac{2 \pi}{7}\right) + \cos \left(\frac{4 \pi}{7}\right) + \cos \left(\frac{6 \pi}{7}\right)$ is equal to

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