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Math Help - sinus and absolute value

  1. #1
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    Exclamation sinus and absolute value

    Hi everyone, i must deduce this inequality : \forall a \in \mathbb{R} \left| sin(a) \right|  \le \left| a \right| from this one : \forall a \in \mathbb{R}+, -a\le sin(a) \le a
    Thank you
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  2. #2
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    Re: sinus and absolute value

    Quote Originally Posted by Groundeleven View Post
    Hi everyone, i must deduce this inequality : \forall a \in \mathbb{R} \left| sin(a) \right|  \le \left| a \right| from this one : \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 \forall a \in \mathbb{R}^+, -a\le sin(a) \le a and from that you need to prove \forall a \in \mathbb{R}^+ \left| sin(a) \right|  \le \left| a \right|~?

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

    This is a standard theorem: -|a|\le b\le |a|\text{ if and only if }|b|\le |a|~.
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    Re: sinus and absolute value

    Plato, I read this in exactly the opposite- the OP knows -a\le sin(x)\le a and must deduce |sin(a)\le |a|. Of course, the definition of |a| is "|a|= a if a\ge 0 and |a|= -a if x< 0". That should be sufficient.
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  4. #4
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    Re: sinus and absolute value

    Quote Originally Posted by HallsofIvy View Post
    Plato, I read this in exactly the opposite- the OP knows -a\le sin(x)\le a and must deduce |sin(a)\le |a|. Of course, the definition of |a| is "|a|= a if a\ge 0 and |a|= -a if x< 0". That should be sufficient.
    How is that opposite from what I posted?
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