While solving some of limits questions I noticed a very remarkable formula popping up which I think is pretty interesting. But I am not able to prove the formula with general values of $n$. The formula goes as follows
$$\begin{align}\phi(0)\sin (a+nh)&-\phi (1)\sin [a+(n-1)h] \\ &+\phi (2)\sin [a+(n-2)h]\\ &-\cdots \\ &+(-1)^n\sin (a) \\ &= \begin{cases} -2^n\sin^n\left(\frac h2\right)\sin \left(a+\frac {nh}{2}\right) , & \text{if $n=4k-2$ $\forall k\in N$} \\[2ex] \phantom{-}2^n\sin^n\left(\frac h2\right)\sin \left(a+\frac {nh}{2}\right) , & \text{if $n=4k\phantom{+0\;}$ $\forall k\in N$ } \\[2ex] -2^n\sin^n\left(\frac h2\right)\cos \left(a+\frac {nh}{2}\right) , & \text{if $n=4k-1$ $\forall k\in N$} \\[2ex] \phantom{-}2^n\sin^n\left(\frac h2\right)\cos \left(a+\frac {nh}{2}\right) , & \text{if $n=4k+1$ $\forall k\in N$} \\[2ex] \end{cases} \end{align}$$
Where $\phi(m)=\binom {n}{m}$ and $n\ge m$, and $n, m\in Z$ and $n\gt 1$
Any resources like PDF or book name are also welcome
Edit:
I used the complex numbers as per the hint but still I am not able to expand it . As per the hint we need to find $$\Im (e^{ia}(e^{ih}-1)^n) =\Im(e^{ia}[(\cos (h) -1)+i\sin h]^n)$$ But I am not able to use the De Moivre's theorem here as it is. Should there be any algebraic manipulations further?