If $m$ and $p$ are positive integers and $m \geq p$, then show that $${m \choose 0}+{m \choose p}+{m \choose 2p}+{m \choose 3p}+\cdots$$ has value $${2^m \over p}\left(1+\sum_{k=1}^{\left \lfloor {p-1 \over 2} \right \rfloor}{{\left( \cos{k\pi \over p}\right)}^m \cos{mk\pi \over p}}\right)$$
I have no idea how to approach this problem. I cannot find a way to express combinatorial arguments using trigonometric functions. It'd be great if someone can give me any hints.
With $\zeta=e^{2\pi\mathrm i/p}$ as Gerry suggested,
\begin{align} \sum_{l=0}^{\left\lfloor\frac mp\right\rfloor}\binom m{lp} &=\sum_{j=0}^m\binom mj\frac1p\sum_{k=0}^{p-1}\zeta^{jk}\\ &=\frac1p\sum_{k=0}^{p-1}\sum_{j=0}^m\binom mj\zeta^{jk}\\ &=\frac1p\sum_{k=0}^{p-1}\left(1+\xi^k\right)^m\\ &=\frac{2^m}p\sum_{k=0}^{p-1}\left(\cos\frac{k\pi}p\right)^m\mathrm e^{mk\pi\mathrm i/p}\\ &=\frac{2^m}p\left(1+2\sum_{k=1}^{\left\lfloor\frac{p-1}2\right\rfloor}\left(\cos\frac{k\pi}p\right)^m\cos\frac{mk\pi}p\right)\;, \end{align}
so I think you were missing a factor of $2$ before the sum.