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Can anyone give me a hint about how to approach this?

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$|\arctan x| \leq \frac{\pi}{4}$ then $$\frac{\pi}{4}\le arctanx\le\frac{\pi}{4}$$ w have $$-\arctan2\le-1\le x\le1\le \arctan2$$ $\to$$$-2\le tanx\le2$$$$-2\le \frac{sinx}{cosx}\le2$$$$-2cosx\le sinx\le2cosx$$

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How do you get to $$-\arctan2\le-1\le x\le1\le \arctan2$$? –  kkaploon Mar 5 '13 at 19:04
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@kkaploon:notice $tan(\frac{\pi}{4})\le tan(arctanx)\le tan(\frac{\pi}{4})$ $\arctan2=1.107148718\ge 1$ and $tan(arctanx)=x$ –  Maisam Hedyelloo Mar 5 '13 at 19:11
    
Thank you very much! :) –  kkaploon Mar 5 '13 at 19:28
    
@ kkaploon:your welcome –  Maisam Hedyelloo Mar 5 '13 at 19:32
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$|\arctan x| \leq \frac{\pi}{4}$ implies $-1\leq x \leq 1$.

Now if you convert $1$ radian into degrees, you will get around $57.3^0$ which is less than $60^0$ and so $tan 1 \leq tan (60^0)=\sqrt{3}\leq 2$ and in this interval cos(x) is positive.

Therefore $|tan(x)|\leq 2$ and from here your result follows

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Why convert into degrees? Why not just say $1\leq \frac{\pi}{3}?$ –  L. F. Mar 5 '13 at 18:18
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