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anushkaudeshan
 4 years ago
(sinasin2a+sin3asin6a+sin4asin13a)/(sinacos2a+sin3asin6a+sin4acos13a)= tan9a
anushkaudeshan
 4 years ago
(sinasin2a+sin3asin6a+sin4asin13a)/(sinacos2a+sin3asin6a+sin4acos13a)= tan9a

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kushashwa23
 4 years ago
Best ResponseYou've already chosen the best response.2post it through equation forum please

lalaly
 4 years ago
Best ResponseYou've already chosen the best response.3\[sinAsin2A = \frac{1}{2}(\cos A  \cos 3A)\] \[\sin3Asin6A = \frac{1}{2}(\cos3A  \cos9A)\] \[\sin4Asin13A = \frac{1}{2}(\cos9A  \cos17A)\] \[sinAcos2A = \frac{1}{2}(sinA+\sin3A) \]\[\sin3Acos6A = \frac{1}{2}(\sin3A+\sin9A)\] \[\sin4Acos13A = \frac{1}{2}(\sin9A+\sin17A)\] \[(sinAsin2A + \sin3Asin6A + \sin4Asin13A) / (sinAcos2A + \sin3Acos6A + \sin4Acos13A) * 2 / 2\] \[= (\cos A  \cos 3A + \cos3A  \cos9A + \cos9A  \cos17A) / (sinA+\sin3A \sin3A+\sin9A \sin9A+\sin17A)\] \[= (\cos A  \cos 17A) / (\sin 17A\sin A)\] \[= 2 \sin 9A \sin 8A / 2 \cos 9A \sin 8A\] (cancel 2 and sin 8A) \[= \sin 9A / \cos 9A\] \[= \tan 9A\]
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