<s>[INST] How to go from $2\cos 2x+1$ to $\frac{\sin 3x}{ \sin x} $ In an exercice I stumbled upon this equation.

$\sin 3x(2\cos 2x+1)=\frac{\sin^23x}{\sin x}$

Or somewhat simplified :

$2\cos 2x+1 = \frac{\sin 3x}{ \sin x} $

I managed to  work out the right to the left , using a lot of formula's (starting with rewriting it as $\frac{\sin (2x + x)}{\sin x }$   and although it would be possible to reverse the steps to go from  left to right, I have the feeling that I am missing something trivial here (just started out to relearn trigonometry to help my kid out...), it's not obvious for me to see how to startout from left to right...
 [/INST] There's nothing wrong with what you did:
$$\frac{\sin(2x+x)}{\sin x} = \frac{\sin(2x)\cos x +\cos(2x)\sin x}{\sin x} $$
$$= \frac{2 \sin(x) \cos (x) \cos (x) + (2 \cos^2 x-1)\sin(x)}{\sin x}$$
$$= 2 \cos^2 x + 2 \cos^2 x - 1$$
$$= 2(2 \cos^2 x - 1) + 1 = 2 \cos 2x + 1$$
where $\cos(2x) = \cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = 2 \cos^2 x - 1$, which can be memorised.
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