Definitely tougher than previous one

Say $S= \sum_{n=1}^{K}\frac{\sin(n\theta)}{2^{n}}$ . Now define $C= \sum_{n=1}^{K}\frac{\cos(n\theta)}{2^{n}}$ .

Evaluating $C\,+\,i \cdot S$ , $C\,+\,i S\,=\,\ \large \sum_{n=1}^{K} \left( \frac{\cos(n\theta)}{2^{n}}\,+\, i \cdot \frac{\sin(n\theta)}{2^{n}} \right) \,=\, \large \sum_{n=1}^{K} \left( \frac{cis(\theta)}{2} \right)^{n}$ $\implies C\,+\,i S\,=\, \frac{cis^{K+1}(\theta)\,-\,cis(\theta) \cdot 2^{K}}{z \cdot 2^{K}\,-\,2^{K+1}}$ After simplification this gives, $C\,+\,i S\,=\, \frac{\cos(K\theta)\,-\,2 \cdot \cos((K+1)\theta)\,-\,2^{K}\,+\,2^{K+1}\cos(\theta)}{2^{K}(5\,-\,4 \cdot \cos(\theta))}\,+\, i \cdot \frac{2^{K+1} \cdot \sin(\theta)\,+\, \sin(K\theta)\,-\,2 \cdot \sin((K+1)\theta)}{2^{K}(5\,-\,4 \cdot \cos(\theta))}$ On equatting the imaginary part, we get :- $\color{#3D99F6}{ S\,=\, \large \sum_{n=1}^{K} \frac{\sin(n\theta)}{2^{n}}\,=\, \frac{2^{K+1} \cdot \sin(\theta) \,+\, \sin(K\theta) \,-\,2 \cdot \sin((K+1)\theta)}{2^{K}(5\,-\,4 \cdot \cos(\theta))} }$ Furthermore, it can also be seen that :- $\color{#D61F06}{C\,=\, \large \sum_{n=1}^{K} \frac{\cos(n\theta)}{2^{n}}\,=\, \frac{\cos(K\theta)\,-\,2 \cdot \cos((K+1)\theta)\,-\,2^{K}\,+\,2^{K+1}\cos(\theta)}{2^{K}(5\,-\,4 \cdot \cos(\theta))} }$