$|e^{-ina}-e^{-inb}|=2\sin(n\theta_0)$ How can I show that $|e^{-ina}-e^{-inb}|=2\sin(n\theta_0)$ for some $\theta_0 \in[a,b]$? I am trying this post Exercise 9, Chapter 2 of Stein's Fourier Analysis. Showing that a fourier series does not converge absolutely but converges conditionally. and was very quickly lost by this fact among others. Is this some application of the mean value theorem that I am not spotting? How can I see this?

You have \begin{align} |e^{-ina}-e^{-inb}|^2 &=(\cos (-na) - \cos (-nb))^2+(\sin (-nb)-\sin (-na))^2\\\ \ \\\ &=2-2\cos(na)\cos(nb)-2\sin(na)\sin(nb)\\\ \ \\\ &=2-2\cos n(a-b)\\\ \ \\\ &=2-2\cos 2\tfrac{n(a-b)}2\\\ \ \\\ &=2-2\left(\cos^2\tfrac{n(a-b)}2-\sin^2\tfrac{n(a-b)}2\right)\\\ \ \\\ &=2-2\left(1-2\sin^2\tfrac{n(a-b)}2 \right)\\\ \ \\\ &=4\sin^2\tfrac{n(a-b)}2. \end{align} So $$ |e^{-ina}-e^{-inb}|=2\left|\sin\tfrac{n(a-b)}2\right| $$