+86 Given constants a, b, and c, let \(\alpha\) and \(\beta\) be solutions to the equations \(a \cos \theta + b \sin \theta = c\) which do not differ by a multiple of \(2\pi\). Express \(\cos^2 \left( \frac{\alpha - \beta}{2} \right)\) in terms of a, b, and c.
+9673 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 \(\cos^2 \dfrac{\alpha - \beta}2 = \dfrac12 (1+1) = 1\).
