I need help to prove that:
$2\cos^2(x)=1+\cos(2x)$.
I know that $\cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$, but I don't know how to get to this step without memorizing this, can you possibly draw some sort of triangle to get this information? I only know the proof that $\cos^2(x)+\sin^2(x)=1$ by using the Pythagoras theorem on a right-angled triangle.
First, you have these identities: $$\cos({\pi \over 2}+x)=-\sin(x),\ \sin({\pi \over 2}+x)=\cos(x)$$
Now consider two right angled triangles glued on their opposites, making a big triangle. With that, do some labeling on the triangles as shown in the picture attached, you can find the areas of the triangles. 
(The area of a triangle is ${1 \over 2}AB\sin(C)$, with $C$ being the included angle in between $A$ and $B$.)
Note that the area of the big triangle is the sum of the areas of the small triangles. $$\therefore {1 \over 2}ae\sin(x+y)={1 \over 2}ac\sin(x)+{1 \over 2}ce\sin(y)$$ $$\therefore \sin(x+y)={{ac}\over {ae}}\sin(x)+{{ce}\over {ae}}\sin(y)$$ $$\because {{ac}\over {ae}}={{c}\over {e}}=\cos(y),\ {{ce}\over {ae}}={{c}\over {a}}=\cos(x)$$ $$\therefore \sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$$ Now, you replace $x$ by ${\pi \over 2} +x$. $$\therefore \sin({\pi \over 2} +x+y)=\cos(y)\sin({\pi \over 2} +x)+\cos({\pi \over 2} +x)\sin(y)$$ With the identities on top, you will get $$\cos(x+y)=\cos(y)\cos(x)-\sin(x)\sin(y)$$ Using this formula and also ${\cos}^2(x)+{\sin}^2(x)=1$, you can get the required double angle for $cosine$. The identities in the post can be deduced: From observation of a single right angled triangle, $$\cos({\pi \over 2}-x)=\sin(x),\ \cos({\pi \over 2}-x)=\sin(x)$$ Replace $x$ by $-x$ and note that $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$, ($\because$ $sine$ and $cosine$ are odd and even functions respectively) you will have the identities. :)
In $cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)$ put $\beta=\alpha$. Then $cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha)=\cos^2(\alpha)-(1-\cos^2(\alpha))=2\cos^2(\alpha)-1.$.
