how to simplify this trigonometric expression without using double angle identities?

[davidtrinh]

I am going to write down a math problem below.

(ex) Without using double angle identities, simplify 2cos(3x)*sinx

my attempt:

2cos(3x)*sinx
=2cos(x+2x)*sinx
=2sinx*cosx*cos(2x)-2sinx*sinx*sin(2x)
=2sinx*cosx*cos(2x)-2(sinx)^2*sin(2x)
=2sinx*cosx*(cosx*cosx-sinx*sinx)-2sin(2x)+2(cosx)^2*sin(2x)
=2sinx*(cosx)^3-2(sinx)^3*cosx-2sin(2x)+2(cosx)^2*sin(2x)

My strategy makes the original expression very messy. I am really stuck. The answer is sin(4x)-sin(2x).

Can someone explain how to do the problem without double angle identities? Thanks a lot.

 

[ksdhart2]

Well, the fact that you're just spinning your wheels and getting nowhere suggest that your tactic may not be the best one. So, let's go back to the drawing board and see if we can find a better one. Do you know the trig identities for turning a product into a sum? Specifically, I think this one will help you:

$\displaystyle cos(x) \cdot sin(y) = \dfrac{1}{2} \cdot \left( sin(x + y) - sin(x-y) \right)$

Try applying that to your problem. Where does that lead you?