How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way. EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:

Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$

I use the fact that $\cos(2x)=2\cos^2(x)-1,  \sin(2x)=2\sin(x)\cos(x)$
(1) LHS = $\frac{\cos(x)-2\cos^2(x)+1}{\sin(x)(1+2\cos(x))}$
(2) Thus it would suffice to simply prove that $\frac{\cos(x)-2\cos^2(x)+1}{1+2\cos(x)}=1-\cos(x)$
(3) Then I just used simple algebra by letting $u = \cos(x)$ then factorising and simplifying.
(4) Since that equals $1-\cos(x)$ then the LHS = $\frac{1-\cos(x)}{\sin(x)} = $ RHS.
Firstly, on the practice exam, we pretty much only had maximum 2-2.5 minutes to prove this, and this took me some trial and error figuring out which double angle formula to use for cos(2x).
This probably took me 5 minutes just experimenting, and on the final exam there is no way I can spend that long.
What is the better way to do this?
EDIT: I also proved it by multiplying the numerator and denominator by $1-\cos(x)$, since I saw it on the RHS. This worked a lot better, but is that a legitimate proof?

Using compound angle formula, we have
$$
\begin{aligned}
\frac{\cos x-\cos 2 x}{\sin x+\sin 2 x} & =\frac{2 \sin \frac{3 x}{2} \sin \frac{x}{2}}{2 \sin \frac{3 x}{2} \cos \frac{x}{2}} \\
& =\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& =\frac{1-\cos x}{\sin x}
\end{aligned}
$$
