Hi everyone, i must deduce this inequality : $\displaystyle \forall a \in \mathbb{R} \left| sin(a) \right| \le \left| a \right|$ from this one : $\displaystyle \forall a \in \mathbb{R}+, -a\le sin(a) \le a$

Thank you :)

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- Sep 3rd 2012, 05:13 AMGroundelevensinus and absolute value
Hi everyone, i must deduce this inequality : $\displaystyle \forall a \in \mathbb{R} \left| sin(a) \right| \le \left| a \right|$ from this one : $\displaystyle \forall a \in \mathbb{R}+, -a\le sin(a) \le a$

Thank you :) - Sep 3rd 2012, 06:23 AMPlatoRe: sinus and absolute value
I am confused as to the exact question.

Are you saying that you know $\displaystyle \forall a \in \mathbb{R}^+, -a\le sin(a) \le a$ and from that you need to prove $\displaystyle \forall a \in \mathbb{R}^+ \left| sin(a) \right| \le \left| a \right|~?$

If so surely you understand that if $\displaystyle a>0\text{ and }-a\le b\le a\text{ then }|b|\le |a|$.

This is a standard theorem: $\displaystyle -|a|\le b\le |a|\text{ if and only if }|b|\le |a|~.$ - Sep 3rd 2012, 06:30 AMHallsofIvyRe: sinus and absolute value
Plato, I read this in exactly the

**opposite**- the OP knows $\displaystyle -a\le sin(x)\le a$ and must deduce |sin(a)\le |a|. Of course, the**definition**of |a| is "|a|= a if $\displaystyle a\ge 0$ and $\displaystyle |a|= -a$ if x< 0". That should be sufficient. - Sep 3rd 2012, 06:37 AMPlatoRe: sinus and absolute value