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Math Help - Help on Verifying identities

  1. #1
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    Help on Verifying identities

    I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:

    sin 5x = 5 sin x - 20 sin^3 x + 16 sin ^ 5 y

    sin 2x + sin 2y + sin 2z = 4 sin x sin y sin z

    sin^2 x + sin ^2 (y-x) + 2 sin x cos y sin (y-x) = sin^2 y


    I'll try to edit this when I come up with answers. Just tips or hints on what identities I should use will do. Thanks for those who will help!
    Last edited by CaptainBlack; September 14th 2008 at 06:10 AM.
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  2. #2
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    Quote Originally Posted by azuresonata View Post
    I was doing an advance reading on trigonometry (identities, specifically) when I encountered these problems on verification:

    sin 5x = 5 sin x - 20 sin^3 x + 16 sin ^ 5 y

    sin 2x + sin 2y + sin 2z = 4 sin x sin y sin z

    sin^2 x + sin ^2 (y-x) + 2 sin x cos y sin (y-x) = sin^2 y


    I'll try to edit this when I come up with answers. Just tips or hints on what identities I should use will do. Thanks for those who will help!

    The second one is not true:

    \sin (2x) + \sin (2y) + \sin (2z) = 4 \sin (x) \sin (y) \sin (z)

    To see this put y=z=0 when the right hand side is identically zero, but the left hand side is not.

    Try the same trick on the third, put y=0, and the first by putting x=0

    RonL
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  3. #3
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    Is substitutuon really okay when answering identities. My aim is to prove this using the trigonometric identities.
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  4. #4
    Super Member Matt Westwood's Avatar
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    Good question.

    It's perfectly okay for proving something is not an identity, but to prove something is an identity you'd have to slot every possible value in to see if it worked. Which is (generally) impossible.
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    Super Member Matt Westwood's Avatar
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    ... but anyway, back to your original question.

    I'd prove the first one by using de Moivre's formula:

    (\cos x + i \sin x)^n = \cos (nx) + i \sin (nx)

    Substitute n=5, multiply out the LHS, simplify out all the algebra (replacing every instance of i^2 with -1, then equating the real and imaginary parts.

    If you haven't investigated complex numbers yet, then perhaps now would be a good time to get into them. Come on in, the water's lovely.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Matt Westwood View Post
    ... but anyway, back to your original question.

    I'd prove the first one by using de Moivre's formula:

    (\cos x + i \sin x)^n = \cos (nx) + i \sin (nx)

    Substitute n=5, multiply out the LHS, simplify out all the algebra (replacing every instance of i^2 with -1, then equating the real and imaginary parts.

    If you haven't investigated complex numbers yet, then perhaps now would be a good time to get into them. Come on in, the water's lovely.
    Unless there is a typo the first is not an identity.

    RonL
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  7. #7
    Super Member Matt Westwood's Avatar
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    Yes, I was taking it for granted that it was a typo.
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