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anushkaudeshan

  • 3 years ago

(sinasin2a+sin3asin6a+sin4asin13a)/(sinacos2a+sin3asin6a+sin4acos13a)= tan9a

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  1. kushashwa23
    • 3 years ago
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    post it through equation forum please

  2. lalaly
    • 3 years ago
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    \[sinAsin2A = \frac{1}{2}(\cos A - \cos 3A)\] \[\sin3Asin6A = \frac{1}{2}(\cos3A - \cos9A)\] \[\sin4Asin13A = \frac{1}{2}(\cos9A - \cos17A)\] \[sinAcos2A = \frac{1}{2}(-sinA+\sin3A) \]\[\sin3Acos6A = \frac{1}{2}(-\sin3A+\sin9A)\] \[\sin4Acos13A = \frac{1}{2}(-\sin9A+\sin17A)\] \[(sinAsin2A + \sin3Asin6A + \sin4Asin13A) / (sinAcos2A + \sin3Acos6A + \sin4Acos13A) * 2 / 2\] \[= (\cos A - \cos 3A + \cos3A - \cos9A + \cos9A - \cos17A) / (-sinA+\sin3A -\sin3A+\sin9A -\sin9A+\sin17A)\] \[= (\cos A - \cos 17A) / (\sin 17A-\sin A)\] \[= 2 \sin 9A \sin 8A / 2 \cos 9A \sin 8A\] (cancel 2 and sin 8A) \[= \sin 9A / \cos 9A\] \[= \tan 9A\]

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