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Math Help - Composite argument property

  1. #1
    Member wiseguy's Avatar
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    Cool Composite argument property

    del
    Last edited by wiseguy; August 6th 2010 at 11:29 AM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    In the "little prince" there is a line: "Draw me a sheep!"

    Now, instead of using composite argument property for cosine and instead drawing me something.... draw to yourself a unit circle.
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  3. #3
    Member wiseguy's Avatar
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    I understand that cos(90-x)=sin, I just have no idea how to interwind it with the composite argument property
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Well, let's try it.

    cos(90 - x) = cos(90)cos(x) + sin(90)sin(x)

    But cos(90) = 0 and sin(90) = 1.

    So?
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Deleted.
    Last edited by Unknown008; July 26th 2010 at 10:50 AM. Reason: dup =(
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by wiseguy View Post
    "Use the composite argument property for cosine to prove that cos (90 -) = sin "


    I'm aware the composite argument property of cosine is cos(A − B) = cos A cos B + sin A sin B - how would you use that to show cos (90 -) = sin ?
    You do realise that your formatting is messed up, right?

    From context, you mean

    Given that \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B

    prove that \cos(90^\circ - \theta) = \sin\theta.

    So let A = 90^\circ and B = \theta and write it out..
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