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3.Trigonometrical Ratios, Functions and Identities
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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
A
$-\frac{1}{2}$
B
$-1$
C
$-\frac{1}{3}$
D
$-\frac{1}{4}$
(JEE MAIN-2022)
Solution
$\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$
$=\frac{\sin \left(3 \times \frac{\pi}{7}\right)}{\sin \frac{\pi}{7}} \times \cos \left(\frac{\frac{2 \pi}{7}+\frac{6 \pi}{7}}{2}\right)$
$=\frac{2 \sin \left(\frac{3 \pi}{7}\right)}{2 \sin \frac{\pi}{7}} \times \cos \left(\frac{4 \pi}{7}\right)$
$=\frac{\sin \left(\frac{7 \pi}{7}\right)+\sin \left(\frac{-\pi}{7}\right)}{2 \sin \frac{\pi}{7}}$
$=\frac{-\sin \frac{\pi}{7}}{2 \sin \frac{\pi}{7}}$
$=-\frac{1}{2}$
Standard 11
Mathematics
