參數方程式

已知

$a\sin \theta + b\cos \theta\,$ 之最大值為 $\sqrt {a^2 + b^2 }\,$
$a\sin \theta + b\cos \theta\,$ 之最小值為 $- \sqrt {a^2 + b^2 }\,$
亦即： $- \sqrt {a^2 + b^2 } \le a\sin \theta + b\cos \theta \le \sqrt {a^2 + b^2 }\,$

$f(\theta ) = a\sin \theta + b\cos \theta\,$
$\Rightarrow f'(\theta ) = a\cos \theta - b\sin \theta$
$\Rightarrow f''(\theta ) = - a\sin \theta - b\cos \theta$
令 $f'(\theta ) = 0\,$
$\Rightarrow a\cos \theta - b\sin \theta = 0\,$
$\Rightarrow a\cos \theta = b\sin \theta \,$
$\Rightarrow {{\sin \theta } \over {\cos \theta }} = {a \over b}$
$\Rightarrow \begin{cases} {\tan \theta = {a \over b}} \\ {\tan \theta = {{ - a} \over { - b}}} \end{cases}$

當
1. $\Rightarrow \begin{cases} {\sin \theta = {a \over {\sqrt {a^2 + b^2 } }}} \\ {\cos \theta = {b \over {\sqrt {a^2 + b^2 } }}} \end{cases}$
2. 上式代入 $f(\theta ) = a\sin \theta + b\cos \theta\,$ ，得
3. $f(\theta ) = a{a \over {\sqrt {a^2 + b^2 } }} + b{b \over {\sqrt {a^2 + b^2 } }}$
4. $\Rightarrow f(\theta ) = {{a^2 } \over {\sqrt {a^2 + b^2 } }} + {{b^2 } \over {\sqrt {a^2 + b^2 } }} = {{a^2 + b^2 } \over {\sqrt {a^2 + b^2 } }}$
5. $\Rightarrow f(\theta ) = {{a^2 + b^2 } \over {\sqrt {a^2 + b^2 } }} \cdot {{\sqrt {a^2 + b^2 } } \over {\sqrt {a^2 + b^2 } }} = {{(a^2 + b^2 )\sqrt {a^2 + b^2 } } \over {a^2 + b^2 }}$
6. $\Rightarrow f(\theta ) = \sqrt {a^2 + b^2 }$
7. 將 $\begin{cases} {\sin \theta = {a \over {\sqrt {a^2 + b^2 } }}} \\ {\cos \theta = {b \over {\sqrt {a^2 + b^2 } }}} \end{cases}\,$ 代入 $f''(\theta ) = - a\sin \theta - b\cos \theta\,$ ，得
8. $f''(\theta ) = - a{a \over {\sqrt {a^2 + b^2 } }} - b{b \over {\sqrt {a^2 + b^2 } }}\,$
9. $\Rightarrow f''(\theta ) = - {{a^2 } \over {\sqrt {a^2 + b^2 } }} - {{b^2 } \over {\sqrt {a^2 + b^2 } }} < 0\,$ ，所以 $f(\theta )\,$ 為最大值。
同理，當
1. $\begin{cases} {\sin \theta = - {a \over {\sqrt {a^2 + b^2 } }}} \\ {\cos \theta = - {b \over {\sqrt {a^2 + b^2 } }}} \end{cases}$
2. $\Rightarrow f(\theta ) = - \sqrt {a^2 + b^2 }\,$
3. $f''(\theta )>0\,$ ，所以 $f(\theta )\,$ 為最小值。