My Math Forum double angle problem

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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 sin2x-cos2x/ 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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