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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.

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Recall that $$\sin(\theta\pm\phi)=\sin(\theta)\cos(\phi)\pm\sin(\phi)\cos(\theta)$$ From this, we have

$$\begin{align} \sin((n+1)x)-\sin((n-1)x) &= \sin(nx)\cos(x)+\sin(x)\cos(nx)-\sin((n-1)x)\\ &= \sin(nx)\cos(x)+\sin(x)\cos(nx)-(\sin(nx)\cos(x)-\sin(x)\cos(nx))\\ &= \sin(nx)\cos(x)+\sin(x)\cos(nx)-\sin(nx)\cos(x)+\sin(x)\cos(nx)\\ &= \sin(nx)\cos(x)-\sin(nx)\cos(x)+\sin(x)\cos(nx)+\sin(x)\cos(nx)\\ &= \sin(x)\cos(nx)+\sin(x)\cos(nx)\\ &\color{green}{= 2\sin(x)\cos(nx)}\\ \end{align}$$

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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 $y \in \mathbb R$.

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Use $$2\cos X \sin Y = \sin (X + Y) - \sin (X - Y) $$,

$$2 \sin(x) \cos(n x) = \sin(x+ n x) - \sin(nx - x) = \sin((n+1)x) - \sin((n-1) x)$$

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Let apply the following Product-to-sum identity

$$2\sin \theta \cos \varphi = {{\sin(\theta + \varphi) + \sin(\theta - \varphi)} }$$

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\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*}

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