1 problem found
Let \[ \mathrm{C}_{n}(\theta)=\sum_{k=0}^{n}\cos k\theta \] and let \[ \mathrm{S}_{n}(\theta)=\sum_{k=0}^{n}\sin k\theta, \] where \(n\) is a positive integer and \(0<\theta<2\pi.\) Show that \[ \mathrm{C}_{n}(\theta)=\frac{\cos(\tfrac{1}{2}n\theta)\sin\left(\frac{1}{2}(n+1)\theta\right)}{\sin(\frac{1}{2}\theta)}, \] and obtain the corresponding expression for \(\mathrm{S}_{n}(\theta)\). Hence, or otherwise, show that for \(0<\theta<2\pi,\) \[ \left|\mathrm{C}_{n}(\theta)-\frac{1}{2}\right|\leqslant\frac{1}{2\sin(\frac{1}{2}\theta)}. \]
Solution: \begin{align*} && C_n(\theta) &= \sum_{k=0}^n \cos k \theta \\ &&&= \textrm{Re} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta}-1}{e^{i\theta}-1}\right)\\ &&&= \textrm{Re} \left ( \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{i(n+1)\theta/2}-e^{-i(n+1)\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}\right)\\ &&&= \textrm{Re} \left ( e^{in\theta/2}\frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Re} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right)\\ \\ && S_n(\theta) &= \sum_{k=0}^n \sin k \theta \\ &&&= \textrm{Im} \left ( \sum_{k=0}^n \exp (ik \theta)\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\textrm{Im} \left ( e^{in\theta/2}\right)\\ &&&= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\sin\left ( \frac12n\theta\right)\\ \\ && C_n(\theta) - \frac12 &= \frac{\sin \left ( (n+1)\theta/2 \right)}{\sin \left ( \theta/2 \right)}\cos \left ( \frac12n\theta\right) - \frac12 \\ &&&= \frac{2\sin \left ( (n+1)\theta/2 \right)\cos\left ( n\theta/2 \right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( (n+1)\theta/2-n\theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (n+1)\theta/2+n\theta/2\right)+\sin\left ( \theta/2\right)-\sin (\theta/2)}{2 \sin (\theta/2)} \\ &&&= \frac{\sin\left ( (2n+1)\theta/2\right)}{2 \sin (\theta/2)} \leqslant\frac{1}{2 \sin (\theta/2)} \\ \end{align*}