Sum of this series.

Question

I tried manipulating it to get it into a binomial expansion of two known terms, but i seemingly failed. Please help me out. $$S=\displaystyle\sum_{r=0}^{12} \binom{12}{r} \cos \frac {r\pi}{6}$$

Answer 1

$\displaystyle\cos\frac{r\pi}6=$Real $(e^{\dfrac{ir\pi}6})$

So, $\displaystyle\sum_{r=0}^{12}\binom{12}r\cos\frac{r\pi}6$=Real $\displaystyle\left(\sum_{r=0}^{12}\binom{12}re^{\dfrac{ir\pi}6}\right)$

Now $\displaystyle\sum_{r=0}^{12}\binom{12}re^{\dfrac{ir\pi}6}=\sum_{r=0}^{12}\binom{12}r(e^{\dfrac{i\pi}6})^r=(1+e^{\dfrac{i\pi}6})^{12}$

and $\displaystyle1+e^{i(2x)}=1+\cos2x+i\sin2x=2\cos x(\cos x+i\sin x)=2e^{ix}\cos x$

Answer 2

Hint: $$\cos\frac{r\pi}{6} = \text{Real}\left(e^{\dfrac{i\pi}{6}}\right)^r$$