# sinus and absolute value

• Sep 3rd 2012, 05:13 AM
Groundeleven
sinus 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 AM
Plato
Re: sinus and absolute value
Quote:

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

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 AM
HallsofIvy
Re: 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 AM
Plato
Re: sinus and absolute value
Quote:

Originally Posted by HallsofIvy
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.

How is that opposite from what I posted?