1. Apr 12, 2009 #1

   fiziksfun

   

   can someone please show me why sinxcos2x = (sin3x - sinx)*0.5
   
   I've been working on it for thirty minutes

    

   fiziksfun, Apr 12, 2009
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3. Apr 12, 2009 #2

   praharmitra

   

   Re: sinxcos2x
   
   can u expand sin(3x) ?? and cos(2x)
   
   just simplify both sides..and u'll get it..

    

   praharmitra, Apr 12, 2009
4. Apr 12, 2009 #3

   qntty

   

   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

   qntty, Apr 12, 2009

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