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anonymous
 4 years ago
sin8x=8sinxcosxcos2xcos4x
proving trig identity?
anonymous
 4 years ago
sin8x=8sinxcosxcos2xcos4x proving trig identity?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sin8x=8sinxcosxcos2xcos4x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have LS: sin8x = sin(4x+4x) = 2sin4xcos4x then i have no idea.. should i just keep on expanding? or is there another way of solving this?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1repeat the process: sin4x=2sin2xcos2x sin2x=2sinxcosx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just keep doing that? :/ would that even get simplified?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ \sin 8x = 2 \sin 4x \cos 4x = 4\sin 2x \cos 2x \cos 4x = 8 \sin x \cos x \cos 2x \cos 4x \]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1not sure if you want to call it 'simplified', but it is what you were looking for

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh. it is! ahaha thanks.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In general, \[ \cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A \cdots \cos 2^{n1}A = \frac { \sin 2^n A}{ 2^n \sin A } \] \[ \implies 2^n \sin A \cdot \cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A \cdots \cos 2^{n1}A= \sin 2^n A\]
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