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November 19th, 2017, 09:59 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 product-to-sum formulas and the 2sin is throwing me off. Write 2sin(3x)cos(x) as a sum |
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November 19th, 2017, 10:17 PM | #2 |
Newbie Joined: Oct 2017 From: Redlands, CA Posts: 15 Thanks: 0 |
Nevermind.. I was overthinking again. The 2 goes away because in product-to-sum you multiply by 1/2.
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November 20th, 2017, 07:26 AM | #3 |
Math Team Joined: Jan 2015 From: Alabama Posts: 3,110 Thanks: 855 |
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). |
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November 20th, 2017, 08:02 AM | #4 |
Global Moderator Joined: Dec 2006 Posts: 18,852 Thanks: 1570 |
The relevant product-to-sum formula is given as $2\sin A\cos B = \sin(A + B) + \sin(A - B)$ in the book I use.
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November 21st, 2017, 08:32 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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