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anonymous

  • 5 years ago

sin 4x = 4 sin x cos³x-4sin³x cos x. Show as an identity.

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  1. anonymous
    • 5 years ago
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    2sin2xcos2x = 4sinxcosx(cos^2 x - sin^2 x)

  2. anonymous
    • 5 years ago
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    thats sin4x

  3. anonymous
    • 5 years ago
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    skipped a few steps there , I see what you have done, but to someone asking the question it is unlikely they will be able to follow

  4. anonymous
    • 5 years ago
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    i assume they know basic trig ratios

  5. anonymous
    • 5 years ago
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    Have difficulty doing these, could you help with the steps so I can try and understand

  6. anonymous
    • 5 years ago
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    sin(2a) = 2sin(a)cos(x)

  7. anonymous
    • 5 years ago
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    cos(2a) = cos^2 (a) - sin^2 (a)

  8. anonymous
    • 5 years ago
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    remember them

  9. anonymous
    • 5 years ago
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    now sin(4x) = sin( 2 * 2x) = 2sin(2x)cos(2x) by applying the first formula once

  10. anonymous
    • 5 years ago
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    hard to memorize, I see those in my notes. Thanks for helping

  11. anonymous
    • 5 years ago
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    now we can apply the first formula again to sin(2x) , to get 2sin(x)cos(x) and we apply the second formula to the cos(2x) term to get cos^2 (x) - sin^2 (x)

  12. anonymous
    • 5 years ago
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    so sin(4x) = 2 * ( 2sin(x)cos(x) ) ( cos^2 (x) - sin^2(x) )

  13. anonymous
    • 5 years ago
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    = 4sin(x)cos(x) ( cos^2(x) - sin^2(x)) , then expand

  14. anonymous
    • 5 years ago
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    thank you!

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