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Question Number 110269 by Khanacademy last updated on 28/Aug/20

Commented by bemath last updated on 28/Aug/20

$$-\frac{1}{2}$$

Commented by Khanacademy last updated on 28/Aug/20

$$?$$

Commented by john santu last updated on 28/Aug/20

$$ithinkthisquestionis$$$$\Rightarrow \cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{6\pi }{7}$$

Commented by bemath last updated on 28/Aug/20

$$if\cos \frac{2\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{6\pi }{7}=p$$$$2p\sin \frac{2\pi }{7}=\sin \frac{4\pi }{7}.\cos \frac{4\pi }{7}.\cos \frac{6\pi }{7}$$$$4p\sin \frac{2\pi }{7}=\sin \frac{8\pi }{7}.\cos \frac{6\pi }{7}$$$$8p\sin \frac{2\pi }{7}=\sin \frac{14\pi }{7}+\sin \frac{2\pi }{7}$$$$[\sin \frac{14\pi }{7}=0]\Rightarrow p=\frac{1}{8}$$

Answered by Dwaipayan Shikari last updated on 28/Aug/20

$$cos2\theta +cos4\theta +cos6\theta \theta =\frac{\pi }{7}7\theta =\pi$$$$\frac{1}{2sin\theta }(sin3\theta -sin\theta +sin5\theta -sin3\theta +sin7\theta -sin5\theta )$$$$\frac{1}{2sin\theta }(sin7\theta -sin\theta )=-\frac{sin\theta }{2sin\theta }=-\frac{1}{2}(sin7\theta =0)$$

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