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Proving it

  1. Oct 9, 2006 #1
    I am trying to prove this equation:

    Sin 6x Cos 4x + Cos 4x sin 2x =
    Cos 2x tan 8x
    Tan 8x​
    does anyone have any idea??
     
  2. jcsd
  3. Oct 9, 2006 #2
    [tex] \sin 6x\cos 4x = \frac{1}{2}(\sin 10x + \sin 2x) [/tex]
    [tex] \cos 4x\sin 2x = \frac{1}{2}(\sin 6x - \sin 2x) [/tex].

    So we get [tex] \frac{1}{2}(\sin 10x + \sin 6x) [/tex]

    or [tex] \sin 8x\cos 2x [/tex].

    The second expression is not right. Just plug in some values.
     
    Last edited: Oct 9, 2006
  4. Oct 9, 2006 #3
    yes I will tray to slove it

    before some month I one like this and I see new law
    tan(a)+tan(b)
    tab(a)-tan(b)

    I discoverd new law juist in my opinion
     
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