Forgot password?
 Register account
View 1399|Reply 5

[不等式] 2个三角函数不等式

[Copy link]

107

Threads

225

Posts

2905

Credits

Credits
2905

Show all posts

facebooker Posted 2021-4-10 12:26 |Read mode
$0<x<\frac{\pi}{2}$, pro:

$\frac{\sin x}{x}>\sqrt[3]{\cos x}$

$e^{\sin x}>x+1$

682

Threads

110K

Posts

910K

Credits

Credits
90973
QQ

Show all posts

kuing Posted 2021-4-10 14:50
第一个:《撸题集》P.1000 题目 6.10.81.

682

Threads

110K

Posts

910K

Credits

Credits
90973
QQ

Show all posts

kuing Posted 2021-4-10 15:06
第二个:令 `f(x)=\sin x-\ln (x+1)`,其中 `0<x<\pi/2`,求导得
\begin{align*}
f'(x)&=\cos x-\frac 1{x+1},\\
f''(x)&=-\sin x+\frac 1{(x+1)^2},\\
f'''(x)&=-\cos x-\frac 2{(x+1)^3}<0,
\end{align*}所以 `f''(x)\searrow `,又 `f''(0)=1`, `f''(\pi /2)=-1+\frac 1{(\pi /2+1)^2}<0`,所以 `f''(x)` 先正后负,即 `f'(x)\nearrow \searrow `,又 `f'(0)=0`, `f'(\pi /2)=-\frac 1{\pi /2+1}<0`,所以 `f'(x)` 先正后负,即 `f(x)\nearrow \searrow `,又 `f(0)=0`, `f(\pi /2)=1-\ln (\pi /2+1)>1-\ln (3.2/2+1)=1-\ln (2.6)>0`,所以 `f(x)>0`,即得证。

1

Threads

15

Posts

334

Credits

Credits
334
QQ

Show all posts

ZCos666 Posted 2025-3-19 13:39
两个都是切线

768

Threads

4685

Posts

310K

Credits

Credits
35004

Show all posts

isee Posted 2025-3-19 15:23
$\mathrm e^{\sin x}>x+1$ 两边取对数 $\sin x>\ln (1+x)$ 就是某年高考题了.
isee=freeMaths@知乎

768

Threads

4685

Posts

310K

Credits

Credits
35004

Show all posts

isee Posted 2025-3-19 15:27
第一个是大一日常了
\begin{gather*}
\sin^3x>x^3\cos x\\
\Leftarrow \left(x-\frac16x^3\right)^3>x^3\left(1-\frac12x^2+\frac1{24}x^4\right)\\
\iff \frac1{24}x^7\left(1-\frac19x^2\right)>0.
\end{gather*}
isee=freeMaths@知乎

Mobile version|Discuz Math Forum

2025-6-6 10:39 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit