How to do the following trigonometric simplification: $ \frac{1- \cos (3\alpha) }{1- \cos (\alpha)} = (1 + 2\cos (\alpha)^2) $ This is probably a trivial question but I don't understand it. Somehow I can't seem to understand how to simplify this expression:
$$ \frac{1- \cos (3\alpha) }{1- \cos (\alpha)} = (1 + 2\cos (\alpha))^2 $$

I prefer using capital $A$, so bare with me.
We will prove that $$1-\cos3A=(1-\cos A)(1+2\cos A)^2$$
First we will show that $$\cos3A=4\cos^3A-3\cos A$$
Indeed, we know that $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$
Thus, for $x=2A, y=A$ we obtain: $$(*):\cos 3A=\cos 2A\cos A-\sin 2A\sin A$$
but from $(*)$ we see that: $\cos 2A=\cos^2A-\sin^2A$
and we know that $\sin 2A=2\sin A\cos A$
So,$$\cos 3A=(\cos^2A-\sin^2A)\cos A-2\sin^2A\cos A\\=(2\cos^2A-1)\cos A-2(1-\cos^2)\cos A\\= 4\cos^3A-3\cos A$$
Now we have $$1-\cos3A=1-4\cos^3A+3\cos A$$
But $$(1-\cos A)(1+2\cos A)^2=\\(1-\cos A)(\cos^2A+2\cos A+1)=\\1-4\cos^3A+3\cos A$$
and we are done.
