I need to prove that: \begin{align} c_k =&\; \cos(k\!\cdot\!x)\\ c_k :=&\; c_{k-1} +d_{k-1}\\ d_k :=&\; 2d_0\!\cdot\!c_k +d_{k−1}\\ d_0 :=&\; −2\!\cdot\!\sin^2{(x/2)}\\ \end{align}
I've got an explicit formula for $d_k$ which should be:
\begin{align} d_k&=d_o+\sum_{i=1}^k{2\!\cdot\!d_o\!\cdot\!c_i} &&\implies& c_k &=c_{k-1}+ \sum_{i=1}^k{2\!\cdot\!d_o\!\cdot\!c_i} \end{align}
Now I want to do a proof by induction. Assuming that $c_p=\cos(p\!\cdot\!x)$ for every $p<k$. This would get me the following: $$c_k =\cos\left(\left(k-1\right)\!\cdot\!x\right)+\sum_{i=1}^k{2\!\cdot\!d_o\!\cdot\!\cos(p\!\cdot\!x)}$$ Using this formula I found:
\begin{align} \sum_{k=1}^n \cos(kx) & = \frac{\sin\left(\frac{nx}2\right)}{\sin \left(\frac{x}2\right)}\, \cos\left(\frac{(n+1)\,x}2\right) \end{align}
I tried to play around with trigonometric addition formulas but I am getting nowhere.