proof "$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$" I want to proof: 
$cos^2(\frac{x}{2})=\frac{1+cos(x)}{2}$
I have changed the given eqation $1 = cos^2(x)+sin^2(x)$
$\to$ $cos^2(x) = 1- sin^2(x)$
Then another given eqation: 
$cos(2x) = cos^2(x)-sin^2(x)$ $\to$ $-sin^2(x)=cos(2x)-cos^2(x)$
After that I have put those equtions together: 
$$cos^2(x)=1-sin^2(x)$$
$$cos^2(x)=1+cos(2x)-cos^2(x)$$
So I have the 1+ {...} structure. How can I go on ?
Or is that the wrong way to proof the equation?
I would really appreciate some hints. 
 A: Simple.  Combine like terms.
$$\cos^2x=1+\cos2x-\cos^2x$$
add $\cos^2x$ to both sides.
$$2\cos^2x=1+\cos2x$$
divide both sides by $2$.
$$\cos^2x=\frac{1+\cos2x}2$$
Let $x=\frac\theta2$ and you get

$$\cos^2\frac\theta2=\frac{1+\cos\theta}2$$

A: $$\cos(2y)=2\cos^2(y)-1$$
Let $2y=x$,
$$\cos(x )= 2\cos^2\left(\frac{x}{2}\right)-1$$
Hence 
$$\frac{1+\cos(x)}{2}=\frac{1+2\cos^2(\frac{x}{2})-1}{2}=\cos^2\left(\frac{x}{2}\right)$$
A: Usually the easiest way to proof trigonometric identities is to use complex exponentials. Here we have $\cos(x) = \frac{1}{2}(e^{ix}+e^{-ix})$ and thus:
$$ \cos^2(\frac{x}{2})
= \frac{1}{4}(e^{i\frac{x}{2}}+e^{-i\frac{x}{2}})^2
= \frac{1}{4}(e^{ix} + 2+ e^{-ix}) = \frac{1}{2}+\frac{1}{2}\cos(x) $$
A: The simple formula for half angle of cosine is $\cos(x)=\cos^2(x/2)-\sin^2(x/2)$.
Lets go from LHS to RHS in your question. Aplying the formula I mentioned, $\frac{1+cos(x)}{2}$ becomes $\frac{1+\cos^2(x/2)-\sin^2(x/2)}{2}$.
If you rewrite 1as $ \cos^2(x/2)+\sin^2(x/2)$ you will get:
$$\frac{\cos^2(x/2)+\sin^2(x/2)+\cos^2(x/2)-\sin^2(x/2)}{2}$$
$$\frac{2\cos^2(x/2)}{2}$$
$$\cos^2(x/2)$$
