trigonometry - Proving $sin ((n-1/2)phi) + sin(phi/2)=sin({n+1 over 2}phi)$

I am trying to show that

$\sin ((n-1/2)\phi) + \sin(\phi/2)=\sin({n+1 \over 2}\phi)$

I tried to apply $\sin (x-y) = \sin x \cos y - \cos x \sin y$ to it and I got

$\sin ((n-1/2)\phi) + \sin(\phi/2)=\sin \phi / 2 \sin n \phi / 2 + \sin n \phi \cos \phi / 2$

but how can I proceed from there?

Answer

You need the prosthaphaeresis/sum-to-product formulae. Adding the expanded forms of $\sin{(x-y)}$ and $\sin{(x+y)}$, you find
$$\sin{(x-y)} + \sin{(x+y)} = 2\sin{x}\cos{y},$$
and changing variables,

$$\sin{A}+\sin{B} = 2\sin{\tfrac{1}{2}(A+B)}\cos{\tfrac{1}{2}(A-B)}.$$
Then the left-hand side of your identity is
$$\sin{(n-\tfrac{1}{2})\phi} + \sin{\tfrac{1}{2}\phi} = 2\sin{\tfrac{n}{2}\phi} \cos{\tfrac{n+1}{2}\phi},$$
which is not often equal to what you have on the right...