JPow

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

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

Country Boy

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

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

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