# Given $|\arctan x| \leq \frac{\pi}{4}$ prove that $|\sin x| \leq 2\cos x$

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

-

## 2 Answers

$|\arctan x| \leq \frac{\pi}{4}$ then $$\frac{\pi}{4}\le arctanx\le\frac{\pi}{4}$$ w have $$-\arctan2\le-1\le x\le1\le \arctan2$$ $\to$$-2\le tanx\le2$$$$-2\le \frac{sinx}{cosx}\le2$$$$-2cosx\le sinx\le2cosx$$ - 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 - Why convert into degrees? Why not just say$1\leq \frac{\pi}{3}?\$ – L. F. Mar 5 '13 at 18:18