Write 2sin(3x)cos(x) as a sum

JPow · November 19th, 2017, 09:59 PM

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

JPow · November 19th, 2017, 10:17 PM

Nevermind.. I was overthinking again. The 2 goes away because in product-to-sum you multiply by 1/2.

Country Boy · November 20th, 2017, 07:26 AM

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).

skipjack · November 20th, 2017, 08:02 AM

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.

JPow · November 21st, 2017, 08:32 AM

Thanks for the replies. It makes a lot more sense to me now.

November 19th, 2017, 09:59 PM	#1
JPow 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
JPow 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
Country Boy Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902	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). Thanks from JPow
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November 20th, 2017, 08:02 AM	#4
skipjack Global Moderator Joined: Dec 2006 Posts: 20,978 Thanks: 2229	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. Thanks from JPow
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November 21st, 2017, 08:32 AM	#5
JPow 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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