Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question

2 Answers 2

up vote 1 down vote accepted

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

share|cite|improve this answer
How do you get to $$-\arctan2\le-1\le x\le1\le \arctan2$$? – kkaploon Mar 5 '13 at 19:04
@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

$|\arctan x| \leq \frac{\pi}{4}$ 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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.