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November 19th, 2017, 10:59 PM  #1 
Newbie Joined: Oct 2017 From: Redlands, CA Posts: 15 Thanks: 0  Write 2sin(3x)cos(x) as a sum
Please help me out with this. I'm new to the producttosum formulas and the 2sin is throwing me off. Write 2sin(3x)cos(x) as a sum 
November 19th, 2017, 11:17 PM  #2 
Newbie Joined: Oct 2017 From: Redlands, CA Posts: 15 Thanks: 0 
Nevermind.. I was overthinking again. The 2 goes away because in producttosum you multiply by 1/2.

November 20th, 2017, 08:26 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
The obvious thing to have done would be to ignore the "2", expand sin(3x)cos(x) and then multiply by 2. For sin(3x)cos(x), I would use the fact that sin(A+ B)= sin(A)cos(B)+ cos(A)sin(B). It is also true, then, that sin(A B)= sin(A)cos(B) cos(A)sin(B). Adding 2sin(A)cos(B)= sin(A+ B)+ sin(A B). Here, 2sin(3x)cos(x)= sin(3x+ x)+ sin(3x x)= sin(4x)+ sin(2x). 
November 20th, 2017, 09:02 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,099 Thanks: 1905 
The relevant producttosum formula is given as $2\sin A\cos B = \sin(A + B) + \sin(A  B)$ in the book I use.

November 21st, 2017, 09:32 AM  #5 
Newbie Joined: Oct 2017 From: Redlands, CA Posts: 15 Thanks: 0 
Thanks for the replies. It makes a lot more sense to me now.


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2sin3xcosx, sum, write 
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