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Trig identity Q

  1. Aug 16, 2007 #1

    T-7

    User Avatar

    Could someone tell me which trig identity this chap (in a Physics textbook) is using to turn

    sin(Kx)sin(2Kx)

    into

    sin^2(3Kx/2) - sin^2(Kx/2)

    Cheers.
     
  2. jcsd
  3. Aug 16, 2007 #2
    [tex]\sin{kx}\sin{2kx}[/tex]

    [tex]=\frac{\cos{(-kx)}-\cos{(3kx)}}{2}[/tex]

    [tex]=\frac{\cos{(kx)}}{2} - \frac{\cos{(3kx)}}{2} [/tex]

    [tex]=\frac{-1 + \cos{(kx)}}{2} - \frac{-1 + \cos{(3kx)}}{2}[/tex]

    [tex]=\sin^2{(3kx/2) - \sin^2{(kx/2)}[/tex]
     
  4. Aug 16, 2007 #3

    T-7

    User Avatar

    Ah... Thank you very much.
     
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