showing that cos(y)-cos(x)=2*sin((x+y)/2)*sin((x-y)/2) how can one show that $\cos(y)-cos(x)=2sin(\frac{x-y}{2})sin(\frac{x+y}{2})$ by just using trigonometrical equalities like additional theorems?
I was trying to show it by rewriting $cos(y)=cos(0.5y+0.5y)$ and then using the additional theorems, but somehow I do not have the result, that is needed.
Starting on the righthandside wasn't successfull either.
Thanks for helping
 A: I think the easier way to do it is by working with the right hand side, instead of the left hand side, making use of the addition formula for sine:
$$\begin{eqnarray}2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right) &=& 2\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{-y}{2}\right)+\cos\left(\frac{x}{2}\right)\sin\left(\frac{-y}{2}\right)\right)\\
&&\times\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right)+\cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)\end{eqnarray}$$
Since cosine is even and sine is odd, we can simplify this to
$$2\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right)-\cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)\\\left(\sin\left(\frac{x}{2}\right)\cos\left(\frac{y}{2}\right)+\cos\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\right)$$
This is just the difference of two squares, so we get
$$2\left(\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{y}{2}\right)-\cos^2\left(\frac{x}{2}\right)\sin^2\left(\frac{y}{2}\right)\right).$$
Since $\sin^2x+\cos^2x=1$, we can rewrite this slightly as
$$2\left(\sin^2\left(\frac{x}{2}\right)\left(1-\sin^2\left(\frac{y}{2}\right)\right)-\cos^2\left(\frac{x}{2}\right)\sin^2\left(\frac{y}{2}\right)\right)$$
Expanding, we get
$$2\sin^2\left(\frac{x}{2}\right)-2\left(\sin^2\left(\frac{x}{2}\right)+\cos^2\left(\frac{x}{2}\right)\right)\sin^2\left(\frac{y}{2}\right)$$
We can see that the inner term has the Pythagorean identity so we can replace it with $1$ to get
$$2\sin^2\left(\frac{x}{2}\right)-2\sin^2\left(\frac{y}{2}\right)$$
Since $\cos(2x) = 1-2\sin^2(x)$, the above simplifies to exactly what we want.
