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

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• 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

$${\displaystyle S_{\triangle OKA}<S_{sectKOA}<S_{\triangle OAL}} \tag1$$ where ${\displaystyle S_{sectKOA}}$ — area of sector ${\displaystyle KOA}$

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