1. Verifying trig identities

Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx.

I know I have to use the product to sum identities but im still stuck.
So far I have, sin(3x+3x)+sin(3x-3x)-sin(x+x)-sin(x-x)
What happens to the sin after you subtract 3x-3x and x-x?

2. Originally Posted by kelsey3
Verify the identity: 2cos4xsin2x=2sin3xcos3x-2cosxsinx.

I know I have to use the product to sum identities but im still stuck.
So far I have, sin(3x+3x)+sin(3x-3x)-sin(x+x)-sin(x-x)
What happens to the sin after you subtract 3x-3x and x-x?

$\displaystyle \underbrace{2\sin(3x) \cos(3x)}_{=\sin(6x)} -2 \sin(x) \cos(x)$

$\displaystyle \underbrace{\sin(6x)}_{\sin(2x)\cos(4x)+\sin(4x)\c os(2x)} -2\sin(x)\cos(x)$

$\displaystyle \sin(2x)\cos(4x)+\underbrace{\sin(4x)}_{2\sin(2x)\ cos(2x)}\cos(2x) -2\sin(x)\cos(x)$

$\displaystyle \sin(2x)\cos(4x)+2\sin(2x) \underbrace{\cos^{2}(2x)}_{\frac{1+\cos(4x)}{2}} -\underbrace{2\sin(x)\cos(x)}_{\sin(2x)}$

$\displaystyle \sin(2x)\cos(4x)+\sin(2x) +\sin(2x)\cos(4x)-\sin(2x)$

$\displaystyle 2 \sin(2x) \cos(4x)$

Yeah!!

Im curious if there is a faster way? anyone?

3. Think Soroban got it in the other, double-posted thread.