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Use the mean value theorem to prove that if $\displaystyle ~|x|<\frac{\pi}2 ~,~~ |y|<\frac{\pi}2~,$ then $$|\sin(y) - \sin(x)| \leq |y-x| \leq |\tan(y) - \tan(x)|$$

What I did was using the mean value theorem using $\sin(x)$ as the function to get $$\frac{ \sin(x) - \sin(y) }{ x-y } = \sin'(c) = \cos(c)$$ for some $c$ inbetween $x$ and $y$, and since the value of $\cos(x)$ for every $x$ goes around $1$ and $-1$ $$\left| \frac{ \sin(x) - \sin(y) }{ x-y }\right| \leq 1$$ but I'm stuck here

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  • $\begingroup$ So, you have proved the first inequality. Can you use the same idea to prove the right inequality, using MVT for $\tan$? $\endgroup$ – GReyes May 1 at 2:31
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Note that if $f(x)=\tan{x}$, then by mean value theorem $$\frac{\tan{x}-\tan{y}}{x-y}=\tan'(c)=\sec^2({c})\geq1$$

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  • $\begingroup$ ok so with this I get (using the transitive property of inequalities) |sin(y)−sin(x)|≤|y−x|≤|tan(y)−tan(x)| but where does the hypothesis come in play? how do the x|<π/2,|y|<π/2 relate to his? $\endgroup$ – rorod8 May 1 at 2:40
  • $\begingroup$ in the case of $\tan{x}$ it is necessary to guarantee continuity, for apply the theore $\endgroup$ – AsdrubalBeltran May 1 at 2:46
  • $\begingroup$ ohhhhhhhh, ok, thank you dude $\endgroup$ – rorod8 May 1 at 2:47

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