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Math Help - show that 8sin20sin40sin80 = root3

  1. #1
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    Lightbulb show that 8sin20sin40sin80 = root3

    show that 8sin20sin40sin80 = root3

    how would I do this?

    Thanks.
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  2. #2
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    Hello differentiate
    Quote Originally Posted by differentiate View Post
    show that 8sin20sin40sin80 = root3

    how would I do this?

    Thanks.
    Start with \sin3A = 3\sin A -4\sin^3 A

    Then \sin 120
     = \frac{\sqrt3}{2} = 3\sin40-4\sin^340

    \Rightarrow\sqrt3= 2\sin40(3-4\sin^240)

    =2\sin40(3-2[1-\cos80]), using \sin^2A = \tfrac12(1-\cos2A)

    =2\sin40(1+2\cos80)

    =4\sin40(\tfrac12+\cos80)

    =4\sin40(\cos60+\cos80)

    =8\sin40\cos70\cos10, using \cos A + \cos B=2\cos\tfrac12(A+B)\cos\tfrac12(A-B)

    =8\sin40\sin20\sin80, using \cos A = \sin(90-A)

    Grandad
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