Problem source

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 source

\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*}