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Math Help - Solving Identities

  1. #1
    cdx
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    Question Solving Identities

    Prove.

    <br />
sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)<br />
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cdx View Post
    Prove.

    <br />
sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)<br />
    sneaky, sneaky! this is actually one of the sum to product formulas in disguise!

    anyway, start with the right side and expand using the addition formulas for sine and cosine

    \sin (A + B) = \sin A \cos B + \sin B \cos A

    \cos (A - B) = \cos A \cos B + \sin A \sin B


    and try to simplify to get the left side, which, if you know your double angle formulas, you'd realize is 2 \sin x \cos x + 2 \sin y \cos y
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  3. #3
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    Quote Originally Posted by cdx View Post
    Prove.

    <br />
sin(2x) + sin(2y) = 2sin(x+y)cos(x-y)<br />
    Another way is to use Euler's formula

    \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2},\;\;\;\sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2}

    Replace the  \theta with  (x+y)\;\;\;(x-y) where appropriate on the RHS, expand and use these expressions again, to get the LHS.
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  4. #4
    cdx
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    Another Identity that I don't quite understand.

    <br />
\frac{1-sin2x}{cos2x}<br />
=<br />
\frac{cos2x}{1+sin2x}<br />
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cdx View Post
    Another Identity that I don't quite understand.

    <br />
\frac{1-sin2x}{cos2x}<br />
=<br />
\frac{cos2x}{1+sin2x}<br />
    you can start with either side here and get the other pretty easily.

    say we start with the left: multiply by \frac {1 + \sin 2x}{1 + \sin 2x}

    it should be easy for you to clean up from there in a step or two (think "difference of two squares")


    P.S. you should post new questions in a new thread
    Last edited by Jhevon; January 22nd 2009 at 11:04 AM.
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