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anonymous

  • 4 years ago

sin8x=8sinxcosxcos2xcos4x proving trig identity?

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  1. anonymous
    • 4 years ago
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    \[\sin8x=8sinxcosxcos2xcos4x\]

  2. anonymous
    • 4 years ago
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    i 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?

  3. TuringTest
    • 4 years ago
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    repeat the process: sin4x=2sin2xcos2x sin2x=2sinxcosx

  4. anonymous
    • 4 years ago
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    just keep doing that? :/ would that even get simplified?

  5. anonymous
    • 4 years ago
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    \[ \sin 8x = 2 \sin 4x \cos 4x = 4\sin 2x \cos 2x \cos 4x = 8 \sin x \cos x \cos 2x \cos 4x \]

  6. TuringTest
    • 4 years ago
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    not sure if you want to call it 'simplified', but it is what you were looking for

  7. anonymous
    • 4 years ago
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    oh. it is! ahaha thanks.

  8. anonymous
    • 4 years ago
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    In general, \[ \cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A \cdots \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } \] \[ \implies 2^n \sin A \cdot \cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A \cdots \cos 2^{n-1}A= \sin 2^n A\]

  9. TuringTest
    • 4 years ago
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    nice^

  10. anonymous
    • 4 years ago
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    Thanks :)

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