+0

# Hard Trigonometry Problem

+1
4
1
+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.

May 13, 2024

#1
+9665
-1

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

May 13, 2024