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Math Help - math is a PROBLEM!

  1. #1
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    math is a PROBLEM!

    sinA

    =

    cosA-coA.cos2A-(1/2)sin2A
    sin2A-cosA



    ALSO


    8sinA.cosA.cos2A.cos4A = sin8A

    how do you do these? and can u explain the 2nd one as well ... thanks
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  2. #2
    Super Member wingless's Avatar
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    Use \sin 2x = 2\sin x \cos x and \cos 2x = \cos^2 x - \sin^2 x.
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  3. #3
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    yeah i know that silly, but how i do it. haha
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  4. #4
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    Hello, Jono Mathematician!

    Wingless suggested a few identities:

    . . \sin2\theta \:=\:2\sin\theta\cos\theta\;\;{\color{blue}[1]}
    . . \cos2\theta \:=\:1-2\sin^2\!\theta\;\;{\color{blue}[2]}

    We must find ways to apply them . . .


    \frac{\cos A - \cos A\cos2A - \frac{1}{2}\sin2A}{\sin2A-\cos A} \;=\;\sin A
    The left side is: . \frac{\cos A - \cos A\overbrace{\cos2A}^{[2]} - \frac{1}{2}\overbrace{\sin2A}^{[1]}}{\underbrace{\sin2A}_{[1]}-\cos A} \;= \;\frac{\cos A - \cos A\overbrace{(1 - 2\sin^2\!A)} - \frac{1}{2}\overbrace{(2\sin A\cos A)}}{\underbrace{2\sin A\cos A} - \cos A}

    . . = \;\frac{\cos A - \cos A + 2\sin^2\!A\cos A - \sin A\cos A}{2\sin A\cos A - \cos A} \;=\;\frac{2\sin^2\!A\cos A - \sin A\cos A}{2\sin A\cos A - cos A}

    . . = \;\frac{\sin A\!\cdot\!\cos A\!\cdot\!(2\sin A - 1)}{\cos A\!\cdot\!(2\sin -1)} \;=\; \sin A



    8\sin A \cos A \cos2A\cos4A \:= \:\sin8A
    \text{The left side is: }\;4\underbrace{(2\sin A\cos A)}\cos2A\cos4A

    . . . . . . . . . = \qquad4\sin2A\cos2A\cos4A

    . . . . . . . . . = \;2\underbrace{(2\sin2A\cos2A)}\cos4A

    . . . . . . . . . =\qquad\underbrace{2\sin4A\cos4A}

    . . . . . . . . . = \;\qquad\quad\sin8A

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