# Where should I find the ranges for $\sum_{n=1}^k \sin n$ and other similar trigonometric series?

It can be found that

$$\sum_{n=1}^k \sin n = \frac{\sin\left(\frac{k+1}{2}\right)\sin\left(\frac{k}{2} \right)}{\sin\left(\frac{1}{2}\right)},$$

$$\sum_{n=1}^k \cos n = \frac{\cos\left(\frac{k+1}{2}\right)\sin\left(\frac{k}{2} \right)}{\sin\left(\frac{1}{2}\right)},$$

and through some manipulation of the trigonometric identities, it can be shown that

$$-0.1277\approx\frac{\cos\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2} \right)} \leq \sum_{n=1}^k \sin n \leq \frac{1+\cos\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)} \approx 1.9582,$$

$$-1.5430\approx\frac{-\sin\left(\frac{1}{2}\right)-1}{2\sin\left(\frac{1}{2}\right)} \leq \sum_{n=1}^k \cos n \leq \frac{1-\sin\left(\frac{1}{2} \right)}{2\sin\left(\frac{1}{2}\right)}\approx0.5430,$$

$$-1.0597\approx\frac{\cos\left(\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)-\sqrt{2}}{2\sin\left(\frac{1}{2}\right)} \leq \sum_{n=1}^k(\cos n+\sin n) \leq \frac{\cos\left(\frac{1}{2}\right)-\sin\left(\frac{1}{2}\right)+\sqrt{2}}{2\sin\left(\frac{1}{2}\right)}\approx1.8902,$$

$$-0.0597\approx\frac{\cos\left(\frac{1}{2}\right)+\sin\left(\frac{1}{2}\right)-\sqrt{2}}{2\sin\left(\frac{1}{2}\right)} \leq \sum_{n=1}^k(\sin n-\cos n) \leq \frac{\cos\left(\frac{1}{2}\right)+\sin\left(\frac{1}{2}\right)+\sqrt{2}}{2\sin\left(\frac{1}{2}\right)}\approx2.8902.$$

Can people please tell me where I can find and confirm these results? I check the inequalities on Desmos, and the actual maximum and minimum values are pretty close to the ideal maximum and minimum values.

• I think it is even more interesting to find an integer $k$ that maximizes or minimizes the expressions. – Szeto Sep 3 '18 at 22:22
• I think there isn't a specific value k that maximizes or minimizes the expressions since there are many integers k that make the expressions approach the ideal maximum and minimum values. For example, when k=334,353,422, the series for cos(n) approaches its maximum value. – Larry Sep 3 '18 at 22:41

Note that $$\sin\left(\frac{k+1}{2}\right) \sin\left(\frac{k}{2}\right) = \frac{\cos(1/2) - \cos(k+1/2)}{2} \in \left[ \frac{\cos(1/2)-1}{2}, \frac{\cos(1/2)+1}{2}\right]$$ and $$\cos\left(\frac{k+1}{2}\right) \sin\left(\frac{k}{2}\right) = \frac{\sin(k+1/2)-\sin(1/2)}{2} \in \left[\frac{-1-\sin(1/2)}{2}, \frac{1-\sin(1/2)}{2} \right]$$