Prove that the $\sum_{x=1}^{n} \cos^2(x)$ is bounded

Question

I need to show that $$\sum_{x=1}^{n} \cos^2(x)$$ is bounded above. I know that there's a similar formula for $$\sum_{x=1}^{n} \sin(2x)$$ but I can't seem to find it anywhere.

Answer 1

$$\begin{align}\sum_{k=1}^{n} \cos^2(k) &= \sum_{k=1}^n \frac{\cos(2k)+1}{2}=\frac 12(n+\sum_{k=1}^n\cos(2k))\\ &=\frac 12(n+Re(\sum_{k=1}^ne^{2ik})) = \frac 12(n+Re(e^{2i}\frac{e^{2in}-1}{e^{2i}-1}))\end{align}$$

Note that $\left|Re(e^{2i}\frac{e^{2in}-1}{e^{2i}-1})\right|\leq \frac{2}{|e^{2i}-1|}$, hence $$\sum_{k=1}^{n} \cos^2(k) = \frac n2 + O(1)$$

This implies the sum is not bounded.