$\sin x < x < \tan x$

Show that for $0 < x < \frac{\pi}{2}$ , we have

$\sin x < x < \tan x .$

This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

#Calculus #Proofs

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

On $[0,\pi/2]$ , consider the two functions : $f(x)=\sin x-x$ , and : $g(x)=\tan x -x$ .

$f'(x)=\cos x -1\leq 0$ , which means that $f(x)\leq f(0)=0$ .

And : $g'(x)=\sec^2 x -1\leq 0$ , which means that $g(x)\geq g(0)=0$ .

And since these two function are not constant on any open interval we get sharp inequalities on the open interval $(0,\pi/2)$ , which get us to what we want.

Haroun Meghaichi - 7 years, 2 months ago

Do you know a non-calculus approach to this problem?

Calvin Lin Staff - 7 years, 2 months ago

We can use Maclaurin expansion of sinx.

Sambit Senapati - 7 years, 2 months ago

I can't draw a circle here, the whole idea is for a point $M$ on the first quadrant (and not on $(1,0)$ ) of the unit circle, the arc from $(1,0)$ to $M$ is longer than the distance between $M$ and the x-axis which gives us $\sin x< x$ . A similar approach would be with the other side of the inequality.

Haroun Meghaichi - 7 years, 2 months ago

By looking at the graphs of sinx , tanx and x we can say it is true.

Akash Shah - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

sin⁡x<x<tan⁡x \sin x < x < \tan x sinx<x<tanx

Comments

$\sin x < x < \tan x$