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Squaring sine in inequality?

  1. Sep 26, 2011 #1
    Hello

    I am doing a calculus proof with epsilon-delta and I am trying to say the following:

    -1[itex]\leq[/itex]sin x[itex]\leq1[/itex]

    and now I want to get (sin x )^2 ...so can you just square all sides of the inequality like this:

    (-1)^2[itex]\leq(sin x)^2[/itex][itex]\leq(1)^2[/itex]

    ??

    According to the rule for inequalities, you can do this i think? But obviously sinx squared isnt between 1 and 1?
     
  2. jcsd
  3. Sep 26, 2011 #2
    Well, you obviously can't :-)

    You can only square an inequality if you know that all the expressions in it are positive.
    In this case -1 isn't positive, so...
     
  4. Sep 26, 2011 #3

    micromass

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    No, you can't do this. The rule

    [tex]a\leq b~\Rightarrow~a^2\leq b^2[/tex]

    only holds if [itex]a,b\geq 0[/itex].

    If both [itex]a,b\leq 0[/itex], then we got the reverse rule

    [tex]a\leq b~\Rightarrow b^2\leq a^2[/tex]

    If we have [itex]a\leq 0\leq b[/itex] then all sort of things can happen. It's not possible to find a relation between [itex]a^2[/itex] and [itex]b^2[/itex] just like that.
     
  5. Sep 26, 2011 #4

    Char. Limit

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    Another possible thing to do would be to multiply by [itex]sin(x)[/itex], which is totally viable for [itex]x \in \left[0, \pi\right][/itex].
     
  6. Sep 26, 2011 #5

    LCKurtz

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    Would squaring the inequality
    [tex]0\leq |sin(x)|\leq 1[/tex]
    help you?
     
  7. Sep 26, 2011 #6
    Yup this helps thx!
     
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