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Math Help - identity showing practice

  1. #1
    Super Member
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    identity showing practice

    hi guys, im tackling this question and need abit of guidence.

     <br />
sin2x + sin4x + sin6x = 4cosxcos2xsin3x<br />


    so far i got:

    ( 2sinxcosx)+(4sin2xcos2x)+(6sin3xcos3x)

    not sure where to go after this?

    any help much appreciated
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  2. #2
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    Somewhere in the south....
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    Continue simplifying the double angles (I can see two in the middle part that can be done right now), and see if you can add the three together.
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  3. #3
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    Quote Originally Posted by atac777 View Post
    Continue simplifying the double angles (I can see two in the middle part that can be done right now), and see if you can add the three together.

    okay umm

    2sinxcosx + 2(2sinxcosx) + 3(2sinxcosx)

    then i dont know what to do.. i get confused easy
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  4. #4
    MHF Contributor
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    Quote Originally Posted by jvignacio View Post
    hi guys, im tackling this question and need abit of guidence.

     <br />
sin2x + sin4x + sin6x = 4cosxcos2xsin3x<br />


    so far i got:

    ( 2sinxcosx)+(4sin2xcos2x)+(6sin3xcos3x)

    not sure where to go after this?

    any help much appreciated
    First you made a mistake
    \sin 4x = 2 \sin2x \cos2x and \sin 6x = 2 \sin3x \cos3x

    Now

    \sin 6x = 2 \sin 3x \cos 3x = 2 \sin 3x (\cos 2x \cos x - \sin 2x \sin x)

    \sin 6x = 2 \cos 2x \cos x \sin 3x - 2 \sin 2x \sin x \sin 3x

    The first part is OK (same form as the second member to find)

    The second part is - 2 \sin 2x \sin x \sin 3x = -2 \sin 2x \sin x (\sin 2x \cos x + \sin x \cos 2x)

    - 2 \sin 2x \sin x \sin 3x = -2 \sin^2 2x \sin x \cos x - 2 \sin^2 x \cos 2x \sin 2x

    - 2 \sin 2x \sin x \sin 3x = -2 (1 - \cos^2 2x) \sin x \cos x - 2 (1 - \cos^2 x) \cos 2x \sin 2x

    Expanding and using 2 \sin x \cos x = \sin 2x and 2 \sin2x \cos2x = \sin 4x

    - 2 \sin 2x \sin x \sin 3x = - \sin 2x + 2 \cos^2 2x \sin x \cos x - \sin 4x + 2 \cos^2 x \cos 2x \sin 2x

    - 2 \sin 2x \sin x \sin 3x = - \sin 2x - \sin 4x + 2 \cos 2x \cos x (\cos 2x \sin x + \cos x \sin 2x)

    - 2 \sin 2x \sin x \sin 3x = - \sin 2x - \sin 4x + 2 \cos 2x \cos x \sin 3x

    Therefore

    \sin 6x = 2 \cos 2x \cos x \sin 3x - \sin 2x - \sin 4x + 2 \cos 2x \cos x \sin 3x

    \sin 2x + \sin 4x + \sin 6x = 4 \cos x \cos 2x \sin 3x
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