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Math Help - Proving identities

  1. #1
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    Proving identities

    Prove (sin2A-sinA)(1+2cosA)=sin3A
    i used the identity sinA-sinB=2cos(A+B)/2*sin(A-B)/2, but then i got stuck.
    Thanks
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  2. #2
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    Hello, arze!

    This requires Olympic-level gymnastics . . .


    Prove: . (\sin2A-\sin A)(1+2\cos A) \:=\:\sin3A

    (\sin2x - \sin A)(1 + 2\cos A) \;=\;(2\sin A\cos A - \sin A)(1 + 2\cos A)

    . . . . . =\;2\sin A\cos A + 4\sin A\cos^2\!A - \sin A - 2\sin A\cos A

    . . . . . - \;4\sin A\underbrace{\cos^2\!A}_{\searrow} - \sin A
    . . . . . = \;4\sin A(1 - \sin^2\!A) - \sin A

    . . . . . = \;4\sin A - 4\sin^3\!A - \sin A

    . . . . . = \qquad\underbrace{3\sin A} \qquad- \qquad \underbrace{4\sin^3\!A}
    . . . . . = \;\overbrace{2\sin A + \sin A} - \overbrace{2\sin^3\!A - 2\sin^3\!A}

    . . . . . = \;2\sin A-2\sin^3\!A + \sin A - 2\sin^3\!A

    . . . . . = \;2\sin A\underbrace{(1 - \sin^2\!A)} \;+ \;\sin A\underbrace{(1 - 2\sin^2\!A)}
    . . . . . = \quad2\sin A\overbrace{\cos^2\!A} \qquad+ \qquad \sin A\overbrace{\cos2A}

    . . . . . = \;\underbrace{(2\sin A\cos A)}\cos A + \sin A\cos2A
    . . . . . =\qquad \overbrace{\sin2A}\cos A + \sin A\cos2A

    . . . . . = \qquad\qquad\qquad\sin3A

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