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Algebra Difficulty 4.8 AIME Prove it China

Suppose α,β≥0\alpha, \beta \ge 0, α+β≤2π\alpha + \beta \le 2\pi. Then the minimum of sin⁡α+2cos⁡β\sin \alpha + 2 \cos \beta is ______.

Solution

When 0≤α≤π0 \le \alpha \le \pi, sin⁡α+2cos⁡β≥0+2⋅(−1)=−2\sin \alpha + 2 \cos \beta \ge 0 + 2 \cdot (-1) = -2.

When π<α≤2π\pi < \alpha \le 2\pi, there is 0≤β≤2π−α<π0 \le \beta \le 2\pi - \alpha < \pi. At this point, as β\beta gets bigger, cos⁡β\cos \beta gets smaller. Therefore,
sin⁡α+2cos⁡β≥sin⁡α+2cos⁡(2π−α)=sin⁡α+2cos⁡α=5sin⁡(α+φ), \begin{aligned} \sin \alpha + 2 \cos \beta &\ge \sin \alpha + 2 \cos(2\pi - \alpha) \\ &= \sin \alpha + 2 \cos \alpha \\ &= \sqrt{5} \sin(\alpha + \varphi), \end{aligned}
where φ=arcsin⁡255\varphi = \arcsin \frac{2\sqrt{5}}{5}.

When α=3π2−φ\alpha = \frac{3\pi}{2} - \varphi, β=2π−α=π2+φ\beta = 2\pi - \alpha = \frac{\pi}{2} + \varphi, sin⁡α+2cos⁡β\sin \alpha + 2 \cos \beta gets the minimum −5-\sqrt{5}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.