1 problem found
Show that \[ 2\sin \frac12 \theta \, \cos r\theta = \sin\big(r+\frac12\big)\theta - \sin\big(r-\frac12\big)\theta \;. \] Hence, or otherwise, find all solutions of the equation \[ \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta = 0 \;, \] where \(a\) and \(b\) are positive integers with \(a < b-1\,\).
Solution: \begin{align*} && \sin\left(r+\frac12\right)\theta - \sin\left(r-\frac12\right)\theta &= \sin r \theta \cos \tfrac12 \theta+\cos r \theta \sin \tfrac12 \theta- \left (\sin r \theta \cos \tfrac12 \theta-\cos r \theta \sin \tfrac12 \theta \right)\\ &&&= 2 \cos r\theta \sin \tfrac12 \theta \end{align*} \begin{align*} && S &= \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta \\ && 2\sin\tfrac12 \theta S &= \sum_{r=a}^{b-1} 2\sin\tfrac12 \theta \cos r \theta \\ &&&= \sum_{r=a}^{b-1} \left ( \sin\left(r+\frac12\right)\theta - \sin\left(r-\frac12\right)\theta \right) \\ &&&= \sin \left (b-\frac12 \right)\theta - \sin \left (a -\frac12 \right)\theta \\ \Rightarrow && \sin \left (b-\frac12 \right)\theta &= \sin \left (a -\frac12 \right)\theta \\ \end{align*} Case 1: \(A = B + 2n\pi\) \begin{align*} && \left (b-\frac12 \right)\theta &= \left (a -\frac12 \right)\theta + 2n\pi \\ \Rightarrow && (b-a) \theta &= 2n \pi \\ \Rightarrow && \theta &= \frac{2n\pi}{b-a} \end{align*} Case 2: \(A = (2n+1)\pi - B\) \begin{align*} && \left (b-\frac12 \right)\theta &= (2n+1)\pi -\left (a -\frac12 \right)\theta \\ \Rightarrow && (b+a-1) \theta &= (2n+1) \pi \\ \Rightarrow && \theta &= \frac{2n\pi}{b+a-1} \end{align*}