두 직선 사잇각과 최대

$\displaystyle \tan ( \theta _{1} + \theta _{2} ) = \sqrt{3}$ $\displaystyle \therefore \theta _{1} + \theta _{2} = \frac{\pi}{3}$ $\tan \theta _{1} = m$이므로 $\therefore 3 \sin \theta _{1} + 4 \sin \theta _{2}$ $\displaystyle = 3 \sin \theta + 4 \sin \left( \frac{\pi}{3} - \theta _{1} \right)$ $\displaystyle = 3 \sin \theta _{1} + 4 \left( \frac{\sqrt{3}}{2} \cos \theta _{1} - \frac{1}{2} \sin \theta _{1} \right)$ $\displaystyle = \sin \theta _{1} + 2 \sqrt{3} \cos \theta _{1}$ $\displaystyle = \sqrt{13} \sin ( \theta _{1} + \alpha ) \left( \cos \alpha = \frac{1}{\sqrt{3}} , \sin \alpha = \frac{2 \sqrt{3}}{\sqrt{13}} \right)$ 최대가 되는 $\displaystyle \theta _{1} = \frac{\pi}{2} - \alpha$이므로 $\displaystyle m = \tan \theta _{1} = \tan \left( \frac{\pi}{2} - \alpha \right)$ $\displaystyle = \frac{\cos \alpha}{\sin \alpha} = \frac{1}{2 \sqrt{3}} = \frac{\sqrt{3}}{6}$