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_{kâ1}\\
d_0 :=&\; â2\!\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.
Best Answer
The last equation reads $$d_0=-2\sin^2\left(\frac x2\right)=\cos(x)-1.$$
Then eliminating $d$ from the second and the third,
$$d_k=c_{k+1}-c_k,\\ c_{k+1}-2c_k+c_{k-1}=2(\cos(x)-1)c_k.$$
If we assume that $c_k=\cos(kx)$, then after some simplification,
$$\cos(kx+k)-2\cos(kx)+\cos(kx-x)=2\cos(kx)\cos(x)-2\cos(kx)\\=2(\cos(x)-1)\cos(kx).$$
Hence $c_k=\cos(kx)$ is indeed a solution, such that $c_0=1$ (not specified in the OP).
We also have
$$d_k=\cos((k+1)x)-\cos(kx).$$
Now knowing these relations, formally proving by induction isn't a problem anymore.
By the induction hypothesis, $$c_{k-1}=\cos((k-1)x),\\d_{k-1}=\cos(kx)-\cos((k-1)x).$$
Then,
$$\color{green}{c_k}=c_{k-1}+d_{k-1}=\cos((k-1)x)+\cos(kx)-\cos((k-1)x)\color{green}{=\cos(kx)}$$ and $$\color{green}{d_k}=2(\cos(x)-1)c_k+d_{k-1}\\ =2(\cos(x)-1)\cos(kx)+\cos(kx)-\cos((k-1)x)\\ =2\cos(x)\cos(kx)-2\cos(kx)+\cos(kx)-\cos(kx)\cos(x)-\sin(kx)\sin(x)\\ =\cos(kx)\cos(x)-\sin(kx)\sin(x)-\cos(kx)\\ \color{green}{=\cos((k+1)x)-\cos(kx)}.$$
And the base case is
$$\color{green}{c_0}=\cos(0\cdot x)=\color{green}{1},\\ \color{green}{d_0}=\cos(1\cdot x)-\cos(0\cdot x)=\color{green}{\cos(x)-1}.$$