<s>[INST] Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&=2\sin x (1 - \sin^2 x) + \sin x - 2\sin^3 x\\
&= " 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \\
&=3\sin x - 4\sin^3 x"
\end{align}
$$
The part of the problem I'm having trouble with is in quotations.
My question:
is how does $\sin x - 2\sin^3 x = 4\sin^3 x$?
I see it as this $\sin x - 2\sin^3 x = 2\sin^{3-1} x = 4\sin x$.
 [/INST] With a hopefully obvious notation,
$$s_3=s_2c+c_2s=2sc^2+(1-2s^2)s=2s(1-s^2)+(1-2s^2)s=3s-4s^3.$$
</s>