Prove that $ \cos x - \cos y = -2 \sin ( \frac{x-y}{2} ) \sin ( \frac{x+y}{2} ) $ Prove that $ \cos x - \cos y = -2 \sin \left( \frac{x-y}{2} \right) \sin \left( \frac{x+y}{2} \right) $ without knowing cos identity
We don't know that $ \cos0 = 1 $
We don't know that $ \cos^2 x + \sin^2 x = 1 $
I have managed to prove it using the above facts, but just realised that I can't use them. Now I have been going in circles for a while.
Any ideas how to prove this or even approach it?
Thanks !

Remember this trick for life: write $x = a+b$ and $y = a-b$. So: $$\begin{cases} x&= a+b \\ y &= a-b \end{cases} \implies \begin{cases} a &= \frac{x+y}{2}\\ b &= \frac{x-y}{2} \end{cases}.$$ 
Then: $$\begin{align} \cos x - \cos y &= \cos(a+b)-\cos(a-b) \\ &= \cos a \cos b - \sin a \sin b - (\cos a \cos b + \sin a \sin b) \\ &= -2 \sin a \sin b \\ &= -2 \sin \left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2} \right).\end{align}$$
