I have the following problem: Find the closed-form solution of:
$$\sum_{k=-m}^m(-1)^k(\cos(k\omega)-\sin(k\omega))$$
Since $\sin(x)=-sin(-x)$ and $\sin(0)=0$, the sine term will cancel. Furthermore, cosine is even, hence, the summation becomes $1+2\sum_{k=1}^m(-1)^k\cos(k\omega)$
My attempt to compute that sum is as follows:
$$\sum_{k=1}^m(-1)^k\cos(k\omega)=-\cos(\omega)-(-1)^{m+1}\cos((m+1)\omega)+\sum_{k=2}^{m+1}(-1)^k\cos(k\omega)$$
Then
$$\sum_{k=1}^m(-1)^k\cos(k\omega)\Big(1-A\Big)=-\cos(\omega)-(-1)^{m+1}\cos((m+1)\omega)$$
My problem is that I don't know what $A$ should be.
I use this approach, because it usually does the trick for this kind of series. However, maybe here a completely different approach is required. I'd appreciate if someone can help me figure out what $A$ should be, or a different approach to find the closed-form solution.
Thanks
Best Answer
Hint
As $k$ is an integer, $(-1)^k=\cos k\pi$
Now use https://mathworld.wolfram.com/WernerFormulas.html
and finally follow How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?