# Thread: Prove the identity sin 3x = 3 sin - 4 sin^3 x

1. ## Prove the identity sin 3x = 3 sin - 4 sin^3 x

This as far as I get to making the left side the same as the right. I could use the steps I am missing

$\displaystyle sin\, 3x = sin(x+2x)$

$\displaystyle = sin\, x \cos 2x + \cos x \sin 2x$

$\displaystyle =\sin x (\cos^2 x- \sin^2 x) + \cos x(2\sin x \cos x)$

but I get lost from here to get the left and right sides the same.

thank you!

2. ## Re: Prove the identity sin 3x = 3 sin - 4 sin^3 x

Originally Posted by Jonroberts74
This as far as I get to making the left side the same as the right. I could use the steps I am missing

$\displaystyle sin\, 3x = sin(x+2x)$

$\displaystyle = sin\, x \cos 2x + \cos x \sin 2x$

$\displaystyle =\sin x (\cos^2 x- \sin^2 x) + \cos x(2\sin x \cos x)$

but I get lost from here to get the left and right sides the same.

thank you!
Let's multiply out what you have:

\displaystyle \begin{align*}\sin x (\cos^2 x- \sin^2 x) + \cos x(2\sin x \cos x) & = \sin x \cos^2 x - \sin^3 x + 2\sin x \cos^2 x \\ & = 3\sin x \cos^2 x - \sin^3 x \\ & = 3\sin x(1-\sin^2 x) - \sin^3 x \\ & = 3\sin x - 3\sin^3 x - \sin^3 x \\ & = 3\sin x - 4\sin^3 x\end{align*}