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November 20th, 2012, 05:31 PM  #1 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  double angle problem
Hi, here's the question simplify the given expression; sin3x  cos3x sinx cos x just not sure which special product to use here? sin2a=2sinacosa is my guess but I don't know how to accommodate the 3x with that. Anybody able to help? If I take LCD, I end up with sin2xcos2x/ sinxcosx. 
November 20th, 2012, 05:41 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,810 Thanks: 2153 
sin(3x)cos(x)  cos(3x)sin(x)  = sin(3x  x)/((sin2x)/2) = 2 sin(x)cos(x) 
November 20th, 2012, 05:56 PM  #3 
Newbie Joined: Feb 2012 Posts: 6 Thanks: 0  Re: double angle problem
Thanx for the reply. still curious how the cos products turned into cos products? was the sin2a=2sinacosa formula used? 
November 21st, 2012, 09:27 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,810 Thanks: 2153 
There were no cos products that turned into cos products!


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angle, double, problem 
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