Evaluate $\sum^{20}_{r=0} \binom{20}{r}\cos r \theta$ 
Evaluate $$\sum^{20}_{r=0} \binom{20}{r}\cos r \theta$$

To do this my initial thought was to set up an alternative series, $S$, $$S=\sum^{20}_{r=0} \binom{20}{r}\sin r \theta$$ If yoy let the original series be $C$, then $$C+iS = \sum^{20}_{r=0} \binom{20}{r}\cos r \theta + i\sum^{20}_{r=0} \binom{20}{r}\sin r \theta $$ 
Nw, I'm not sure how to get this into a ncie obvious looking form, but i am assuming it is equal to, $$C+iS = 1+20(\cos \theta + i\sin\theta) + \binom{20}{2}(\cos 2\theta + i\sin2\theta) + \cdots (\cos20\theta + i\sin20\theta)$$
$$C+iS = 1+20(\cos \theta + i\sin\theta) + \binom{20}{2}(\cos \theta + i\sin\theta)^2 + \cdots (\cos\theta + i\sin\theta)^{20}$$
$$C+iS = (1+(\cos\theta + i\sin\theta))^{20}$$ This is the point I didn't know where to continue from. All I know is that I want to get to a point where I can just take the real part which would be $C$. I know the answer is $$C=2^{20} \cos^{20}(\frac12\theta) \cos10\theta $$
 A: Use this:
$$1+\cos{\theta}+i\sin{\theta}=1+e^{i\theta}=2\cos{\frac{\theta}{2}}e^{i\frac{\theta}{2}}$$
Here is how to prove it:
$$1+e^{i\theta} =e^{i\frac{\theta}{2}}e^{-i\frac{\theta}{2}}+e^{i\frac{\theta}{2}}e^{i\frac{\theta}{2}}=e^{i\frac{\theta}{2}}(e^{i\frac{\theta}{2}}+e^{-i\frac{\theta}{2}})=2\cos{\frac{\theta}{2}}e^{i\frac{\theta}{2}}$$
A: I think you should use this $\cos \theta = {e^{i \theta}+ e^{-i\theta}\over 2}$ directy. $\sum_\limits{r = 0}^{20}\begin{pmatrix} 20 \\r \end{pmatrix} \cos r \theta = {1\over 2}\sum_\limits{r = 0}^{20} \begin{pmatrix} 20 \\r \end{pmatrix}({e^{\theta}})^r \cdot 1^{20-r} + {1\over 2}\sum_\limits{r = 0}^{20}\begin{pmatrix} 20 \\r \end{pmatrix}(e^{-\theta})^r \cdot1^{20-r} $
A: You did very well. Now
$$
\eqalign{
  & C + i\,S = \sum\limits_r {\left( \matrix{
  20 \cr 
  r \cr}  \right)e^{\,ir\,\theta } }  = \sum\limits_r {\left( \matrix{
  20 \cr 
  r \cr}  \right)\left( {e^{\,i\,\theta } } \right)^{\,r} }  =   \cr 
  &  = \left( {1 + e^{\,i\,\theta } } \right)^{\,20}   \cr 
  & \left| {1 + e^{\,i\,\theta } } \right| = \sqrt {\left( {1 + \cos \theta } \right)^{\,2}  + \sin ^{\,2} \theta }  = \sqrt {2 + 2\cos \theta }  = 2\cos \left( {\theta /2} \right)  \cr 
  & \phi  = \angle \left( {1 + e^{\,i\,\theta } } \right) = \arctan \left( {{{\sin \theta } \over {1 + \cos \theta }}} \right) =   \cr 
  &  = \arctan \left( {{{2\sin \left( {\theta /2} \right)\cos \left( {\theta /2} \right)} \over {1 + \cos ^{\,2} \left( {\theta /2} \right) - \sin ^{\,2} \left( {\theta /2} \right)}}} \right) =   \cr 
  &  = \arctan \left( {{{\sin \left( {\theta /2} \right)} \over {\cos \left( {\theta /2} \right)}}} \right) = \theta /2 \cr} 
$$
and you know how to conclude.
