Understanding how to arrive at the identity $2 \sin(x) \cos(n x) = \sin((n + 1) x) - \sin((n - 1) x)$ Apologies, I'm a bit rusty on my trig. I am trying to manipulate the LHS side of the equation below to arrive at the RHS.
$$2 \sin(x) \cos(n x) = \sin((n + 1) x) - \sin((n - 1) x)$$
I've tried using the double angle formula to rewrite the cos(nx) but I'm not getting anywhere with it.
Thanks in advance.
 A: Recall that
$$\sin(\theta\pm\phi)=\sin(\theta)\cos(\phi)\pm\sin(\phi)\cos(\theta)$$
From this, we have
$$\begin{align}
\sin((n+1)x)-\sin((n-1)x)
&= \sin(nx)\cos(x)+\sin(x)\cos(nx)-\sin((n-1)x)\\
&= \sin(nx)\cos(x)+\sin(x)\cos(nx)-(\sin(nx)\cos(x)-\sin(x)\cos(nx))\\
&= \sin(nx)\cos(x)+\sin(x)\cos(nx)-\sin(nx)\cos(x)+\sin(x)\cos(nx)\\
&= \sin(nx)\cos(x)-\sin(nx)\cos(x)+\sin(x)\cos(nx)+\sin(x)\cos(nx)\\
&= \sin(x)\cos(nx)+\sin(x)\cos(nx)\\
&\color{green}{= 2\sin(x)\cos(nx)}\\
\end{align}$$
A: Write :
\begin{align}
2 \sin x \cos nx & = (\sin x\cos nx - \cos x\sin nx) + (\sin x \cos nx + \cos x \sin nx) \\& =\sin(x - nx) + \sin(x + nx) \\& = \sin((1-n)x) + \sin((n+1)x) \\&= \sin((n+1)x) - \sin((n-1)x)  
\end{align}
Here, the first line adds and subtracts the same term, the second line uses standard trigionometric formulas, and the fourth uses the fact that $\sin(-y) = -\sin(y)$ for all $y \in \mathbb R$.
A: Use $$2\cos X \sin Y =   \sin (X + Y) - \sin (X - Y)  $$,
$$2 \sin(x) \cos(n x) = \sin(x+ n x) - \sin(nx - x) = \sin((n+1)x) - \sin((n-1) x)$$
A: \begin{eqnarray*}
\sin((n+1)x)=\sin nx \cos x +\cos nx \sin x \\
 \sin((n-1)x)=\sin nx \cos x -\cos nx \sin x \\
\end{eqnarray*}
Subtract these equations & we have
\begin{eqnarray*}
2\cos nx \sin x =sin((n+1)x)-\sin((n-1)x).\\
\end{eqnarray*}
