Let $z=\cos\theta + i\sin\theta$. Then we have $z^n=\cos n\theta + i\sin n\theta$. It follows that \begin{align*} &1 + C + iS\\ &= 1 + C_n^1(\cos\theta + i\sin\theta)+C_n^2(\cos 2\theta +i\sin 2\theta)+\cdots + C_n^n(\cos n\theta +i\sin n\theta)\\ &= C_n^0z^0 + C_n^1z+C_n^2z^2 +\cdots + C_n^nz^n\\ &= (1+z)^n \end{align*} Meanwhile, we have \begin{align*} (1+z)^n &= (1+\cos\theta + i\sin\theta)^n\\ &= \Big(2\cos^2\frac{\theta}{2}+2\sin\frac{\theta}{2}\cos\frac{\theta}{2}i\Big)^n\\ &= 2^n\cos^{2n}\frac{\theta}{2}\Big(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\Big)^n\\ &=2^n\cos^{2n}\frac{\theta}{2}\Big(\cos\frac{n\theta}{2}+i\sin\frac{n\theta}{2}\Big) \end{align*} Therefore, we conclude \begin{align*} S&=\boxed{2^n\cos^n\frac{\theta}{2}\sin\frac{n\theta}{2}}\\ C&=\boxed{-1+2^n\cos^n\frac{\theta}{2}\cos\frac{n\theta}{2}} \end{align*}