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I'm reading Calculus (by Apostol). There is an exercise (page 133 n°34 Spanish version) that asks:

Prove that $\sin x<x$ if $0<x<\frac{\pi}{2}$, looking that the area of $\Delta OAP$ is less than the area of circle section $OAP$.

The radius of circle is $1$. So the area of triangle $<$ area of circle section

$ \frac{\cos x \sin x}{2}< \frac{x}{2}$

$2\cos x \sin x <2x$

$\sin 2x<2x$ where $0<x<\frac{\pi}{2}$ then $0<2x< \pi$

if I take $y = 2x$ then $\sin y <y$ if $0<y<\pi$

then $\sin x <x$ if $0<x<\frac{\pi}{2}$ because $(0,\frac{\pi}{2}) \subset (0,\pi)$

I think this has an error. Indeed, I don't know if the last step is correct. need some help.

Martin Sleziak
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Elll
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    MathJax hint: If you put backslashes before sin and cos, they come out in the correct font. \sin gives $\sin$ while sin gives $sin$. – Ross Millikan Jun 03 '15 at 01:06
  • Have a look at some answers to this question: http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1 – Martin Sleziak Nov 22 '15 at 07:19

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He proved $\,\sin y<y\,$ if $\,0<y<\pi\,$ and deduced that $\,\sin x <x\,$ if $\,0<x<\dfrac\pi2$.

Isn't that clear? He who can do the more can do the less…

Bernard
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