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Math Help - Prove The Identity

  1. #1
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    Prove The Identity

    sin8x= (8sinx)(cosx)(cos2x)(cos4x)

    i have been trying the double angle formula on sin8x and not having any luck any help is appreciated!
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by superhume View Post
    sin8x= (8sinx)(cosx)(cos2x)(cos4x)

    i have been trying the double angle formula on sin8x and not having any luck any help is appreciated!
    sin(8x) = sin2(4x)

    you should know that sin2x = 2sinxcosx. Applying this to the above, you have

    sin2(4x) = 2sin(4x).cos(4x)

    again,

    sin(4x) = sin2(2x) = 2 sin(2x)cos(2x), so you have,

    sin(8x) = 2 [ 2 sin(2x)cos(2x)] cos(4x) = 4 sin(2x)cos(2x)cos(4x)

    again put sin2x = 2sinxcosx in the above, and you have your answer
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  3. #3
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    Hello, superhume!

    Prove: . \sin8x \:=\: 8\sin x\cos x\cos2x\cos4x
    We can work it from the right side . . .


    \text{We have: }\:8\sin x\cos x\cos2x\cos4x \;=\;4\underbrace{(2\sin x\cos x)}_{\text{This is }\sin2x}\cos2x\cos4x

    . . . . . . . . . . . . . . . . . . . . . =\;4\sin2x\cos2x\cos4x

    . . . . . . . . . . . . . . . . . . . . . =\;2\underbrace{(2\sin2x\cos2x)}_{\text{This is }\sin4x}\cos4x

    . . . . . . . . . . . . . . . . . . . . . =\;\underbrace{2\sin4x\cos4x}_{\text{This is }\sin8x}

    . . . . . . . . . . . . . . . . . . . . . =\;\sin8x


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We can work it from the left side . . .


    \text{We have: }\;\sin8x \;=\;2\underbrace{\sin4x}\cos4x

    . . . . . . . . =\; 2\overbrace{(2\sin2x\cos2x)}\cos4x

    . . . . . . . . =\;4\underbrace{\sin2x}\cos2x\cos4x

    . . . . . . =\;4\overbrace{(2\sin x\cos x)}\cos2x\cos4x

    . . . . . . =\;8\sin x\cos x\cos2x\cos4x

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