Proving $\theta < \tan\theta$ with geometry [closed]

I'm looking for a simple geometrical method of proving that $$\theta < \tan\theta$$ for $0 < \theta < \frac{\pi}{2}$.

I am able to prove that $\sin\theta < \theta$.

closed as off-topic by Namaste, Brian Borchers, kingW3, Raskolnikov, José Carlos SantosJan 4 '18 at 22:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Brian Borchers, kingW3
If this question can be reworded to fit the rules in the help center, please edit the question.

• I presume you mean, "for all $0 < \theta < \frac{\pi}{2}$"? Yours sounds like you are looking for just one $\theta$ where $\theta < \tan \theta$. – Mike Pierce Dec 31 '17 at 17:38
• I'm curious. How did you prove "geometrically" that $\sin\theta < \theta$? That seems rather odd since $\theta$ is an angle but $\sin\theta$ is a length/measure. – Mike Pierce Dec 31 '17 at 17:42
• @MikePierce: θ is not an angle, but a measure of an angle, i.e. the length of an arc on the unit circle (or if you allow negative values, a curvilinear abscissa on the trigonometric circle). – Bernard Dec 31 '17 at 19:42
• See my related answer here. – dxiv Dec 31 '17 at 21:50
• – Raskolnikov Jan 4 '18 at 19:41 $$S_{\triangle OKA}<S_{sectKOA}<S_{\triangle OAL}} \tag$$ where $S_{sectKOA}$ — area of sector $KOA}$

$$S_{\triangle KOA}={\frac {1}{2}}\cdot |OA|\cdot |KH|={\frac {1}{2}}\cdot |OA|\cdot |OK|\cdot \sin x={\frac {1}{2}}\cdot 1\cdot 1\cdot \sin x={\frac {\sin x}{2}}$$ $$S_{sectKOA}={\frac {1}{2}}R^{2}x={\frac {x}{2}}$$ $$S_{\triangle OAL}={\frac {1}{2}}\cdot |OA|\cdot |LA|={\frac {\mathrm {tan} \,x}{2}}$$ from $\triangle OAL: |LA|={\mathrm {tan}}\,x$

substitute in $(1)$:

$${\frac {\sin x}{2}}<{\frac {x}{2}}<{\frac {{\mathrm {tan}}\,x}{2}}$$

• I'm not familiar with much of the notation you are using. – Mike Pierce Dec 31 '17 at 17:45
• what do you mean? may be "sector"? (en.wikipedia.org/wiki/Circular_sector) – aid78 Dec 31 '17 at 17:48
• Is a sector the region itself, or the area of the region? What is $S_{\triangle\text{stuff}}$? What is $\operatorname{tg}x$? Am I suppose to guess that $R$ is the radius of the circle and $x$ is the angle in question? It's a pain to sift through your notation. – Mike Pierce Dec 31 '17 at 17:53
• sorry I used $tg x$ like $tan x$, now I'll change it – aid78 Dec 31 '17 at 17:56
• $S_{\triangle OKA}$ means the area of triangle $OKA$ – aid78 Dec 31 '17 at 17:59