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November 19th, 2017, 10:59 PM   #1
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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, 11:17 PM   #2
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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, 08:26 AM   #3
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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, 09:02 AM   #4
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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, 09:32 AM   #5
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Thanks for the replies. It makes a lot more sense to me now.
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