$\begin{array}{l}
P = \sin a + \sqrt 3 \cos a = 2.\left( {\dfrac{1}{2}\sin a + \dfrac{{\sqrt 3 }}{2}\cos a} \right) = 2.\left( {\sin \dfrac{\pi }{6}\sin a + \cos \dfrac{\pi }{6}\cos a} \right) = 2\cos \left( {a - \dfrac{\pi }{6}} \right)\\
- 1 \le \cos \left( {a - \dfrac{\pi }{6}} \right) \le 1 \to P \ge 2
\end{array}$
Dấu bằng xảy ra khi và chỉ khi:
$\cos \left( {a - \dfrac{\pi }{6}} \right) = - 1 \to a = k\pi + \dfrac{\pi }{6} = 2k\pi - \dfrac{{5\pi }}{6}$
