What is $\sup\limits_{n \in \mathbb{N}} \sum\limits_{k=0}^n \cos(k)$? For the sake of curiosity, does anyone know what $\sup\limits_{n \in \mathbb{N}} \sum\limits_{k=0}^n \cos(k)$ or $\sup\limits_{n \in \mathbb{N}} \sum\limits_{k=0}^n \sin(k)$ are? It is pretty easy to see that both partial sums are bounded, but I don't even have a clue as to what their respective supremums should be. Definitely something not too large. 
 A: As shown in the link that dezdichado gave in the comments, we can write 
$\displaystyle\sum_{k = 0}^{n}\cos(k) = \sum_{k = 0}^{n}\dfrac{\cos(k)\sin (\tfrac{1}{2})}{\sin (\tfrac{1}{2})} = \sum_{k = 0}^{n}\dfrac{\sin(\tfrac{k+1}{2})-\sin(\tfrac{k-1}{2})}{2\sin (\tfrac{1}{2})}$ 
$= \dfrac{\sin(\tfrac{n+1}{2})-\sin (-\tfrac{1}{2})}{2\sin (\tfrac{1}{2})} = \dfrac{\sin(\tfrac{n+1}{2})+\sin (\tfrac{1}{2})}{2\sin (\tfrac{1}{2})}$
and
$\displaystyle\sum_{k = 0}^{n}\sin(k) = \sum_{k = 0}^{n}\dfrac{\sin(k)\sin (\tfrac{1}{2})}{\sin (\tfrac{1}{2})} = \sum_{k = 0}^{n}\dfrac{\cos(\tfrac{k-1}{2})-\sin(\tfrac{k+1}{2})}{2\sin (\tfrac{1}{2})}$ 
$= \dfrac{\cos(-\tfrac{1}{2})-\cos(\tfrac{n+1}{2})}{2\sin (\tfrac{1}{2})} = \dfrac{\cos(\tfrac{1}{2})-\cos(\tfrac{n+1}{2})}{2\sin (\tfrac{1}{2})}$.
You can prove that $\displaystyle\sup_{n \in \mathbb{N}}\sin(\tfrac{n+1}{2}) = 1$ and $\displaystyle\inf_{n \in \mathbb{N}}\cos(\tfrac{n+1}{2}) = -1$ using arguments similar to what is discussed here. 
Combining those results gives $\displaystyle\sup_{n \in \mathbb{N}}\sum_{k = 0}^{n}\cos(k) = \dfrac{1}{2\sin(\tfrac{1}{2})}+\dfrac{1}{2}$ and $\displaystyle\sup_{n \in \mathbb{N}}\sum_{k = 0}^{n}\sin(k) = \dfrac{\cos(\tfrac{1}{2})+1}{2\sin (\tfrac{1}{2})}$.
