삼각함수 합 최댓값

$\sin \alpha + \cos \alpha + \sin \beta + \cos \beta$ $\displaystyle = \sqrt{2} \sin \left( \alpha + \frac{\pi}{4} \right) + \sqrt{2} \sin \left( \beta + \frac{\pi}{4} \right)$ $\displaystyle = \sqrt{2} \left\{ \sin \left( \alpha + \frac{\pi}{4} \right) + \sin \left( \beta + \frac{\pi}{4} \right) \right\}$ $\displaystyle = 2 \sqrt{2} \sin \left( \frac{\alpha + \beta}{2} + \frac{\pi}{4} \right) \cos \left( \frac{\alpha - \beta}{2} \right)$ $\displaystyle = 2 \sqrt{2} \sin \left( \frac{\alpha + \beta}{2} + \frac{\pi}{4} \right) \cos \frac{\pi}{6}$ $\displaystyle = \sqrt{6} \sin \left( \frac{\alpha + \beta}{2} + \frac{\pi}{4} \right)$ 그러므로 $\displaystyle \sin \left( \frac{\alpha + \beta}{2} + \frac{\pi}{4} \right) = 1$일 때 최댓값 $\displaystyle M = \sqrt{6}$이고, $M ^{2} = 6$ [참고] $\displaystyle \alpha = \frac{5}{12} \pi$이고 $\displaystyle \beta = \frac{\pi}{12}$일 때 두 식 $\displaystyle \frac{\alpha + \beta}{2} = \frac{\pi}{4}$과 $\displaystyle \frac{\alpha - \beta}{2} = \frac{\pi}{6}$를 만족시킨다.