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?
 A: $$\frac{\cos x-\cos2x}{\sin x+\sin 2x } = \frac{1-\cos x }{\sin x}\iff \sin x\cos x-\sin x\cos2x=\sin x-\sin x\cos x+$$
$$+\sin2x-\sin2x\cos x\iff \color{red}{\sin x\cos 2x}+\sin x+\sin2x-\color{red}{\sin2x\cos x}-2\sin x\cos x=0\iff$$
$$\color{red}{\sin(-x)}+\sin x=0$$
and we're done by the double implications all through (and assuming the first expression is well defined, of course)
Check all the cancellations are correct and check all the trigonometric identities used above.
Another way: We begin with the left side, again: assuming it is well defined
$$\frac{\cos x-\cos2x}{\sin x+\sin2x}\stackrel{\cos2x=2\cos^2x-1\\\sin2x=2\sin x\cos x}=\frac{\cos x-2\cos^2x+1}{\sin x(1+2\cos x)}\stackrel{-2t^2+t+1=-(2t+1)(t-1)}=$$
$$=\require{cancel}-\frac{\cancel{(2\cos x+1)}(\cos x-1)}{\sin x\cancel{(1+2\cos x)}}\stackrel{\cdot\frac{\cos x+1}{\cos x+1}}=-\frac{\overbrace{(\cos^2x-1)}^{=-\sin^2x}}{\sin x(\cos x+1)}=$$
$$=-\frac{(-\sin x)}{(\cos x+1)}=\frac{\sin x}{\cos x+1}$$
Finally, we show that last right side equals the right side of the original equation:
$$\frac{\sin x}{\cos x+1}\cdot\frac{\cos x-1}{\cos x-1}=\frac{\sin x(\cos x-1)}{\underbrace{\cos^2x-1}_{=-\sin^2x}}=-\frac{\cos x-1}{\sin x}=\frac{1-\cos x}{\sin x}$$
A: Here's a trigonographic demonstration:

A: In the domain of your equality the following expression is equivalent:
$\sin(x)(\cos(x)-\cos(2x))=(1-\cos(x))(\sin(x)+\sin(2x)) \iff \\
2\sin(x)\cos(x)= \sin(x)+\sin(2x)+ \sin(x)\cos(2x)-\cos(x)\sin(2x) \iff \\ 2\sin(x)\cos(x)= \sin(x)+\sin(2x)+\sin(x-2x) \iff 2\sin(x)\cos(x)=\sin(2x) $
This guarantees your result because you move from one expression to other by adding or subtracting equivalent values.
A: By the duplication formulas, for $\sin(x)\neq 0$ and $\cos(x)\neq -\frac12$
$$\frac{\cos(x)\color{red}{-\cos(2x)}}{\sin(x)\color{blue}{+\sin(2x)}} =\frac{1-\cos(x)}{\sin(x)}$$
$$\iff \frac{\cos(x)\color{red}{+1-2\cos^2(x)}}{\sin(x)\color{blue}{+2\sin(x)\cos(x)}} =\frac{1-\cos(x)}{\sin(x)}$$
$$\iff \frac{1+\cos(x)-2\cos^2(x)}{1+2\cos(x)} =1-\cos(x)$$
$$\iff 1+\cos(x)-2\cos^2(x) =(1-\cos(x))(1+2\cos (x))$$
which is true.
A: If you are familiar with complex number:
Let $z$ be a complex number satisfying $|z|=1$. We want to prove that $z-\frac{1}{z^{2}}=k\left(1-\frac{1}{z}\right)$ where $k$ is real. After some algebraic manipulation, we obtain $k=z+1+\frac{1}{z}$ which is always real.
A: Using compound angle formula, we have
$$
\begin{aligned}
\frac{\cos x-\cos 2 x}{\sin x+\sin 2 x} & =\frac{2 \sin \frac{3 x}{2} \sin \frac{x}{2}}{2 \sin \frac{3 x}{2} \cos \frac{x}{2}} \\
& =\frac{2 \sin ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\
& =\frac{1-\cos x}{\sin x}
\end{aligned}
$$
