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 !
 A: You may use the complex definition of the sin and cos functions derived from the Euler's formula.
$$\begin{array}{l}\cos x = \frac{{{e^{ix}} + {e^{ - ix}}}}{2},\sin x = \frac{{{e^{ix}} - {e^{ - ix}}}}{{2i}};\\ - 2\sin (\frac{{x - y}}{2})\sin (\frac{{x + y}}{2})\\ =  - 2\left( {\frac{{{e^{i(\frac{{x - y}}{2})}} - {e^{ - i(\frac{{x - y}}{2})}}}}{{2i}}} \right)\left( {\frac{{{e^{i(\frac{{x + y}}{2})}} - {e^{ - i(\frac{{x + y}}{2})}}}}{{2i}}} \right)\\ = \frac{1}{2}\left( {{e^{i(\frac{{x - y}}{2})}} - {e^{ - i(\frac{{x - y}}{2})}}} \right)\left( {{e^{i(\frac{{x + y}}{2})}} - {e^{ - i(\frac{{x + y}}{2})}}} \right)\\ = \frac{1}{2}\left( {{e^{ix}} - {e^{ - iy}} - {e^{iy}} + {e^{ - ix}}} \right)\\ = \frac{1}{2}\left( {{e^{ix}} + {e^{ - ix}}} \right) + \frac{1}{2}\left( { - {e^{ - iy}} - {e^{iy}}} \right)\\ = \frac{1}{2}\left( {{e^{ix}} + {e^{ - ix}}} \right) - \frac{1}{2}\left( {{e^{ - iy}} + {e^{iy}}} \right)\\ = \cos x - \cos y\end{array}$$
A: 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: $$\begin{align} \cos x - \cos y &= \cos(a+b)-\cos(a-b) \\ &= \cos a \cos b - \sin a \sin b - (\cos a \cos b + \sin a \sin b) \\ &= -2 \sin a \sin b \\ &= -2 \sin \left(\frac{x+y}{2}\right) \sin\left(\frac{x-y}{2} \right).\end{align}$$
