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Math Help - expression manipulation

  1. #1
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    expression manipulation

    I want to show that (sin2kx / 2sinx) + cos(2k-1)x = sin(2k+2)x / 2sinx

    I presume I could convert the cos expression to sin first. This is more difficult than the trigonometry I normally do.
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  2. #2
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    \sin{[(2k + 2)x]} = \sin{(2kx + 2x)}

     = \sin(2kx)\cos(2x) + \cos(2kx)\sin(2x)

     = \sin(2kx)[1 - 2\sin^2{x}] + 2\sin{x}\cos{x}\cos{(2kx)}

     = \sin(2kx) - 2\sin(2kx)\sin^2{x} + 2\sin{x}\cos{x}\cos(2kx)

    = \sin(2kx) - 2\sin{x}[\sin(2kx)\sin{x} + \cos(2kx)\cos{x}]

     = \sin(2kx) - 2\sin{x}\cos{(2kx - x)}

     = \sin(2kx) - 2\sin{x}\cos[(2k - 1)x].


    Therefore

    \displaystyle{\frac{\sin{[(2k + 2)x]}}{2\sin{x}} = \frac{\sin(2kx) - 2\sin{x}\cos[(2k - 1)x]}{2\sin{x}}}

    \displaystyle{ = \frac{\sin(2kx)}{2\sin{x}} - \cos[(2k-1)x]}
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  3. #3
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    Thanks, that was more work than I thought it would be. You have made a mistake at the 5th line and further. The last + should be -.
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