Hard Trigonometry Problem

By half angle formula, we have 

$\quad \cos^2 \dfrac{\alpha - \beta}2\\ = \dfrac12 (1 + \cos(\alpha - \beta))\\$

Since $\alpha$ and $\beta$ do not differ by a multiple of $2\pi$, that means $\sin \alpha$, $\sin \beta$ are distinct, and $\cos \alpha$, $\cos \beta$ are distinct. 

From the equation, we have

$b\sin \theta = c - a \cos \theta\\ b^2 \sin^2 \theta = c^2 + a^2 \cos^2 \theta - 2ac \cos \theta\\ b^2 - b^2 \cos^2 \theta = c^2 + a^2 \cos^2 \theta - 2ac \cos \theta\\ (a^2 + b^2) \cos^2 \theta - 2ac \cos \theta + (c^2 - b^2) = 0$

Then by Vieta's formula, $\cos \alpha \cos \beta = \dfrac{c^2 - b^2}{a^2 + b^2}$.

Similarly, we have

$a \cos \theta = c - b\sin \theta\\ a^2 \cos^2 \theta = c^2 + b^2 \sin^2 \theta - 2bc \sin \theta\\ a^2 - a^2 \sin^2 \theta = c^2 + b^2 \sin^2 \theta - 2bc \sin \theta\\ (a^2 + b^2)\sin^2 \theta - 2bc \sin \theta + (c^2 - a^2) = 0$

Then $\sin \alpha \sin \beta = \dfrac{c^2 - a^2}{a^2 + b^2}$.

Hence, we have $\cos(\alpha - \beta) = \dfrac{c^2 - a^2 - (c^2 - b^2)}{a^2 + b^2} = 1$, so $\cos^2 \dfrac{\alpha - \beta}2 = \dfrac12 (1+1) = 1$.