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[函数] 求 $f(\theta)=\frac{\sin \theta+3\cos \theta+3}{\cos \theta}$ 的值域

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isee Posted at 2023-1-21 17:29:03 |Read mode
源自知乎提问


:求 $y=f(\theta)=\dfrac{\sin \theta+3\cos \theta+3}{\cos \theta}$ 的值域.





大约还是辅助角公式比较轻快明了 \begin{gather*}
y=\frac{\sin \theta+3\cos \theta+3}{\cos \theta},\\[1ex]
y\cos\theta=\sin \theta+3\cos \theta+3,\\[1ex]
\sin \theta+(3-y)\cos \theta=-3,\\[1ex]
\sqrt{1^2+(3-y)^2}\sin (\theta+\varphi)=-3,\tan\varphi=\frac{3-y}1.
\end{gather*} 由三角函数的有界性,知 \begin{gather*}
\sqrt{1^2+(3-y)^2}\geqslant \left|-3\right|\\[1ex]
\color{blue}{1+(3-y)^2\geqslant 9},\\[1ex]
\Rightarrow y\leqslant 3-2\sqrt 2\; \lor\;y\geqslant 3+2\sqrt 2.
\end{gather*}

isee=freeMaths@知乎

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其妙 Posted at 2023-1-22 23:57:57
\[y = f(\theta ) = \frac{{\sin \theta  + 3\cos \theta  + 3}}{{\cos \theta }} = \frac{{\sin \theta  + 3\cos \theta  + \sqrt {9({{\sin }^2}\theta  + {{\cos }^2}\theta )} }}{{\cos \theta }}\]
(当 $\cos \theta > 0$时)\[ = \frac{{\sin \theta  + 3\cos \theta  + \sqrt {(1 + 8)({{\sin }^2}\theta  + {{\cos }^2}\theta )} }}{{\cos \theta }}\]
\[ \ge \frac{{\sin \theta  + 3\cos \theta  + \sqrt {{{( - \sin \theta  + 2\sqrt 2 \cos \theta )}^2}} }}{{\cos \theta }}\]
\[ \ge \frac{{\sin \theta  + 3\cos \theta  + ( - \sin \theta  + 2\sqrt 2 \cos \theta )}}{{\cos \theta }} = 3 + 2\sqrt 2 \]
同理,当 $\cos \theta < 0$时,
\[y = f(\theta ) = \frac{{\sin \theta  + 3\cos \theta  + \sqrt {(1 + 8)({{\sin }^2}\theta  + {{\cos }^2}\theta )} }}{{\cos \theta }}\]
\[ \le \frac{{\sin \theta  + 3\cos \theta  + \sqrt {{{( - \sin \theta  - 2\sqrt 2 \cos \theta )}^2}} }}{{\cos \theta }}\]
\[ \le \frac{{\sin \theta  + 3\cos \theta  + ( - \sin \theta  - 2\sqrt 2 \cos \theta )}}{{\cos \theta }} = 3 - 2\sqrt 2 \]
综上所述,\[y \le3 - 2\sqrt 2 \]或\[y \ge3 + 2\sqrt 2 \]
妙不可言,不明其妙,不着一字,各释其妙!

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其妙 Posted at 2023-1-22 23:59:09
较麻烦的写个常数柯西
妙不可言,不明其妙,不着一字,各释其妙!

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 Author| isee Posted at 2023-1-23 09:08:46
其妙 发表于 2023-1-22 23:59
较麻烦的写个常数柯西


写这么长公式,不容易~~
isee=freeMaths@知乎

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色k Posted at 2023-1-23 11:08:20 From the mobile phone
isee 发表于 2023-1-23 09:08
写这么长公式,不容易~~
{{\sin }^2}\theta  + {{\cos }^2}\theta 很明显是机器转码😁
好久没见其妙😁
这名字我喜欢

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hejoseph Posted at 2023-1-23 11:46:32
对于 $\theta$ 有范围限制的时候,下面这个方法比较实用:
利用万能代换,令
\[
\sin\theta=\frac{2t}{1+t^2},\cos\theta=\frac{1-t^2}{1+t^2},
\]

\[
\frac{\sin\theta+3\cos\theta+3}{\cos\theta}=\frac{2(3+t)}{1-t^2}=\frac{2}{6-\left((3+t)+\dfrac{8}{3+t}\right)},
\]
后面容易了。

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