We will make use of two trig identities: \[ \cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\ \cos(A) - \cos(B) = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \]
Using these identities gives \[ \begin{align} N(\theta) &= \left(\cos((m-1)\theta) + \cos(m \theta)\right) -\left(\cos((k-1)\theta) +\cos(k \theta)\right) \\ \\ &= 2\cos\left(\frac{2m-1}{2}\theta\right)\cos\left(\frac{1}{2}\theta\right) - 2\cos\left(\frac{2k-1}{2}\theta\right)\cos\left(\frac{1}{2}\theta\right) \\ \\ &= 2\cos\left(\frac{1}{2}\theta\right) \left[\cos\left(\frac{2m-1}{2}\theta\right) - \cos\left(\frac{2k-1}{2}\theta\right) \right] \\ \\ &= -4\cos\left(\frac{1}{2}\theta\right)\sin\left(\frac{m+k-1}{2}\theta\right)\sin\left(\frac{m-k}{2}\theta\right). \end{align} \]
We consider integers w in the form w = 4n, w = 4n+1, w = 4n+2, and w = 4n+3 for n ≥ 1.
If \(\theta = \dfrac{720^\circ}{4n}\), then \[ \begin{align} k &= \left\lceil{\frac{180^\circ}{\theta}}\right\rceil = \left\lceil{\frac{180(4n)}{720}}\right\rceil = \lceil{n}\rceil = n \\ \\ m &= \left\lceil\frac{540^\circ}{\theta}\right\rceil = \left\lceil\frac{540(4n)}{720}\right\rceil = \left\lceil{3n}\right\rceil = 3n \end{align} \]
Hence \[ \sin\left(\frac{m-k}{2}\theta\right) = \sin\left(\frac{2n}{2}\frac{720^\circ}{4n}\right) = \sin(180^\circ) = 0 \Rightarrow N(\theta) = 0. \] If \(\theta = \dfrac{720^\circ}{4n+b}\), where b = 1, 2, or 3, then \[ \begin{align} k &= \left\lceil{\frac{180^\circ}{\theta}}\right\rceil = \left\lceil{\frac{180(4n+b)}{720}}\right\rceil = \left\lceil{{n+\frac{b}{4}}}\right\rceil = n+1 \\ \\ m &= \left\lceil\frac{540^\circ}{\theta}\right\rceil = \left\lceil\frac{540(4n+b)}{720}\right\rceil = \left\lceil{3n+\frac{3b}{4}}\right\rceil = 3n+b \end{align} \] Hence \[ \sin\left(\frac{m+k-1}{2}\theta\right) = \sin\left(\frac{4n+b}{2}\frac{720^\circ}{4n+b}\right) = \sin(360^\circ) = 0 \Rightarrow N(\theta) = 0. \]Here is a graph of the zeros for 48° ≤ θ ≤ 180°.
Zeros \(\dfrac{720^\circ}{w}\) for w = 1, 2, ..., 15
