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Why sinxcos2x = (sin3x - sinx)*0.5

  1. Apr 12, 2009 #1
    can someone please show me why sinxcos2x = (sin3x - sinx)*0.5

    I've been working on it for thirty minutes
  2. jcsd
  3. Apr 12, 2009 #2
    Re: sinxcos2x

    can u expand sin(3x) ?? and cos(2x)

    just simplify both sides..and u'll get it..
  4. Apr 12, 2009 #3
    Re: sinxcos2x

    You should use the identities

    [tex]\cos{2\theta} = 1-2\sin^2{\theta}[/tex]

    [tex]\sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}[/tex]

    Perhaps you were putting these identities in terms of cosines. When possible, stick with as few trig functions as possible and replace the [tex]\cos^2{\theta}[/tex] with [tex]1-\sin^2{\theta}[/tex] or vice versa.
    Last edited: Apr 12, 2009
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