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 $\sin x - 2\sin^3 x = 2\sin^{3-1} x = 4\sin x$.

Observe it's rather:
$$
\begin{align}
\color{blue}{2\sin x} - 2\sin^3 x + \color{blue}{\sin x} - 2\sin^3 x &=\color{blue}{2\sin x} + \color{blue}{\sin x}- 2\sin^3 x  - 2\sin^3 x 
\\\\&=3\sin x - 4\sin^3 x.
\end{align}
$$
