Cosine of the sum of two solutions of trigonometric equation $a\cos \theta + b\sin \theta = c$ Question:
If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that:
$$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$
No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help!

Here's a picture showing angles $\theta$ (at $P$) and $\phi$ (at $Q$) such that
$$a \cos\theta + b \sin\theta = c = a \cos\phi + b \sin \phi$$
The measure of $\angle PAQ$ is the sum of these angles.

Note that $P$ and $Q$ lie on the circle with diameter $\overline{AB}$, and that the diameter bisects $\angle PAQ$. From here, we have many approaches to the final relation; here's one: Clearly,
$$\cos\frac{\theta+\phi}{2} = \frac{a}{d} \qquad\qquad \sin\frac{\theta+\phi}{2} = \frac{b}{d}$$
so that, by the Double-Angle Formulas,
$$\cos(\theta+\phi) = 2\cos^2\frac{\theta+\phi}{2} - 1 = \frac{2a^2-d^2}{d^2} = \frac{2a^2-(a^2+b^2)}{a^2+b^2} = \frac{a^2-b^2}{a^2+b^2}$$
$$\sin(\theta+\phi) = 2 \sin\frac{\theta+\phi}{2}\cos\frac{\theta+\phi}{2} = \frac{2ab}{d^2} = \frac{2ab}{a^2+b^2} $$
