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 $\sin\theta < \theta$.


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  • $\begingroup$ I presume you mean, "for all $0 < \theta < \frac{\pi}{2}$"? Yours sounds like you are looking for just one $\theta$ where $\theta < \tan \theta$. $\endgroup$ – Mike Pierce Dec 31 '17 at 17:38
  • $\begingroup$ 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. $\endgroup$ – Mike Pierce Dec 31 '17 at 17:42
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    $\begingroup$ @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). $\endgroup$ – Bernard Dec 31 '17 at 19:42
  • $\begingroup$ See my related answer here. $\endgroup$ – dxiv Dec 31 '17 at 21:50
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    $\begingroup$ Possible duplicate of Proving $n\sin(\frac{\pi}{n})&lt;\pi&lt;n\tan(\frac{\pi}{n})$ ; obtaining results from it. $\endgroup$ – Raskolnikov Jan 4 '18 at 19:41

enter image description here $${\displaystyle S_{\triangle OKA}<S_{sectKOA}<S_{\triangle OAL}} \tag1$$ 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 $(1)$:

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

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    $\begingroup$ I'm not familiar with much of the notation you are using. $\endgroup$ – Mike Pierce Dec 31 '17 at 17:45
  • $\begingroup$ what do you mean? may be "sector"? (en.wikipedia.org/wiki/Circular_sector) $\endgroup$ – aid78 Dec 31 '17 at 17:48
  • $\begingroup$ 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. $\endgroup$ – Mike Pierce Dec 31 '17 at 17:53
  • $\begingroup$ sorry I used $tg x$ like $tan x$, now I'll change it $\endgroup$ – aid78 Dec 31 '17 at 17:56
  • $\begingroup$ $ S_{\triangle OKA}$ means the area of triangle $OKA$ $\endgroup$ – aid78 Dec 31 '17 at 17:59

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