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Math Help - Simplifying the expression

  1. #1
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    Simplifying the expression

    hey guys having trouble simplying this expression. Any help? thank u

    <br /> <br />
2sin3xcos3xcos5x-(cos^23x-sin^23x)sin5x<br /> <br /> <br />
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  2. #2
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    Use the double angle formulae for sin and cosine.

    (Sorry for my poor help but I'm in a rush.)
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    Use the double angle formulae for sin and cosine.

    (Sorry for my poor help but I'm in a rush.)
    Just to add more:

    \cos{2x} = \cos^2{x} - \sin^2{x}

    \cos{4x} = \cos^2{2x} - \sin^2{2x}
    \ldots

    ==================

    \sin{2x} = 2\sin{x}\cos{x}

    \sin{4x} = 2\sin{2x}\cos{2x}
    \ldots

    Do you see the pattern here?
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  4. #4
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    Quote Originally Posted by Chop Suey View Post
    Just to add more:

    \cos{2x} = \cos^2{x} - \sin^2{x}

    \cos{4x} = \cos^2{2x} - \sin^2{2x}
    \ldots

    ==================

    \sin{2x} = 2\sin{x}\cos{x}

    \sin{4x} = 2\sin{2x}\cos{2x}
    \ldots

    Do you see the pattern here?
    yes yes i see it. so using these double angled formulas i answer this question
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  5. #5
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    Quote Originally Posted by Chop Suey View Post
    Just to add more:

    \cos{2x} = \cos^2{x} - \sin^2{x}

    \cos{4x} = \cos^2{2x} - \sin^2{2x}
    \ldots

    ==================

    \sin{2x} = 2\sin{x}\cos{x}

    \sin{4x} = 2\sin{2x}\cos{2x}
    \ldots

    Do you see the pattern here?
    ok so:
    2sin3xcos3xcos5x-(cos^23x-sin^23x)sin5x

    simplifys to:

    <br />
sin6xcos^2\frac{5}{2}x-sin^2\frac{5}{2}x-(cos6x)(2sin\frac{5}{2}xcos\frac{5}{2}x)<br />

    this correct?
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  6. #6
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    You only needed to change the 2sin3xcos3x and (cos^2(3x) - sin^2(3x)).

    Then, you'll see that this fits the sum identitiy for sine. Recall that:

    \sin{(A+B)} = \sin{A}\cos{B} + \cos{A}\sin{B}
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  7. #7
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    Quote Originally Posted by Chop Suey View Post
    You only needed to change the 2sin3xcos3x and (cos^2(3x) - sin^2(3x)).

    Then, you'll see that this fits the sum identitiy for sine. Recall that:

    \sin{(A+B)} = \sin{A}\cos{B} + \cos{A}\sin{B}
    yes yes thats right.
    so its

    sin6xcos5x-cos6xsin5x

    ?
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  8. #8
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    Quote Originally Posted by jvignacio View Post
    yes yes thats right.
    so its

    sin6xcos5x-cos6xsin5x

    ?
    Yes. Don't doubt yourself.

    And don't forget:
    \sin{(A\pm B)} = \sin{A}\cos{B} \pm \cos{A}\sin{B}
    Last edited by Chop Suey; September 9th 2008 at 05:42 PM.
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  9. #9
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    Quote Originally Posted by Chop Suey View Post
    Yes. Don't doubt yourself.

    And don't forget:
    thanks man, will do. appreciate the help
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