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Math Help - trigo sum/different~do not do~!

  1. #1
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    Post trigo sum/different~do not do~!

    Express product as a sum/different containing only sines or cosines.

    sin4Asin2A=????

    I stuck at beginning.

    THX WHO HELP~!
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  2. #2
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    remember that
    \sin 2A=2\sin A \cos A

    (so obviously \sin 4A = 2\sin2A \cos 2A)

    and  \cos 2A = \cos^2A - \sin^2 A

    make substitutions and simplify
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  3. #3
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    Quote Originally Posted by sanikui View Post
    Express product as a sum/different containing only sines or cosines.

    sin4Asin2A=????

    I stuck at beginning.

    THX WHO HELP~!
    See my reply in this thread for the formulae you need: http://www.mathhelpforum.com/math-he...dentities.html
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  4. #4
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    Wink answer

    sin4Asin2A=sin2(2A)sin2A
    =2sin2Acos2Asin2A
    =2(2sinAcosA)(1-2sin^2A)(2sinAcosA)
    =8sin^2Acos^2A(1-2sin^2A)
    =8sin^2A(1-sin^2A)(1-2sin^2A)
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  5. #5
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    Quote Originally Posted by Mr.Green View Post
    sin4Asin2A=sin2(2A)sin2A
    =2sin2Acos2Asin2A
    =2(2sinAcosA)(1-2sin^2A)(2sinAcosA)
    =8sin^2Acos^2A(1-2sin^2A)
    =8sin^2A(1-sin^2A)(1-2sin^2A)
    No. The answer has to be a sum.

    From the compound angle formulae: cos(x - y) - cos(x + y) = 2 sin(x) sin(y).

    Substitute x = 4A and y = 2A:

    \sin (4A) \sin (2A) = \frac{1}{2} (\cos (2A) - \cos(6A) ).
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  6. #6
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    I got it from Mr.fantastic.
    THx Mr.Green ans Mr.fantastic~!!
    I'm continues do my homework~!!
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