Alright, so the standard way to evaluate $\sum\limits_{k=0}^n\cos(kx)$ and $\sum\limits_{k=0}^n\sin(kx)$, is to respectively take the real and imaginary part of $$\sum_{k=0}^n{\rm e}^{ikx}={\frac {{{\rm e}^{i \left( n+1 \right) x}}-1}{{{\rm e}^{ix}}-1}}$$ $$=\frac12\left({\frac {\cos \left( \left( n+1 \right) x \right) -\cos \left( nx \right) }{\cos \left( x \right) -1}}+1 \right) +\frac12i\left( {\frac {\sin \left( \left( n+1 \right) x \right) -\sin \left( nx \right) -\sin \left( x \right) }{\cos \left( x \right) -1}} \right)$$
While this is relatively simple concept and easy to remember how to derive, is there a way to come up with these results that do not rely on complex numbers? Also, do not give an inductive proof, as that is not particularly constructive, and the form of the sum values is not easy to determine by inspecting.
