# Stuck with trig identities

• April 6th 2008, 03:55 PM
kelsey3
Stuck with 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?
• April 6th 2008, 04:42 PM
Soroban
Hello, kelsey3!

We need this double-angle identity: . $2\sin\theta\cos\theta \:=\:\sin2\theta$

And this sum-to-product identity: . $\sin A - \sin B \:=\:2\cos\left(\frac{A+B}{2}\right)\sin\left(\fra c{A-B}{2}\right)$

Quote:

Verify the identity: . $2\cos4x\sin2x\;=\;2\sin3x\cos3x-2\cos x\sin x$

The right side is: . $\underbrace{2\sin3x\cos3x} - \underbrace{2\sin x\cos x}$

. . . . . . . . . $= \qquad\quad\underbrace{\sin6x \quad\;\;- \;\;\quad \sin 2x}$

. . . . . . . . . $= \quad \overbrace{2\cos\left(\frac{6x+2x}{2}\right)\sin\l eft(\frac{6x-2x}{2}\right)}$

. . . . . . . . . $= \qquad\qquad\quad 2\cos4x\sin 2x$

• April 6th 2008, 04:54 PM
kelsey3
Thank you sooo much!