Using the Mean Value Theorem, prove this inequality.

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Using the Mean Value Theorem, prove that



$|cos^2(b)-cos^2(a)|gt frac14|b-a| $



for all $a,b in (fracpi4,fracpi3)$



So far, I have



Let $f(x)=cos^2(x)$



Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.



Applying the Mean Value Theorem to $f(x)$, we have:



$fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$



$fraccos^2(b)-cos^2(a)b-a= -sin(2x)$



$|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$



Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.



Or am I making a mistake?




real-analysis




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    up vote
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    down vote

    favorite












    Using the Mean Value Theorem, prove that



    $|cos^2(b)-cos^2(a)|gt frac14|b-a| $



    for all $a,b in (fracpi4,fracpi3)$



    So far, I have



    Let $f(x)=cos^2(x)$



    Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.



    Applying the Mean Value Theorem to $f(x)$, we have:



    $fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$



    $fraccos^2(b)-cos^2(a)b-a= -sin(2x)$



    $|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$



    Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.



    Or am I making a mistake?




    real-analysis




    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Using the Mean Value Theorem, prove that



      $|cos^2(b)-cos^2(a)|gt frac14|b-a| $



      for all $a,b in (fracpi4,fracpi3)$



      So far, I have



      Let $f(x)=cos^2(x)$



      Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.



      Applying the Mean Value Theorem to $f(x)$, we have:



      $fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$



      $fraccos^2(b)-cos^2(a)b-a= -sin(2x)$



      $|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$



      Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.



      Or am I making a mistake?




      real-analysis




      share|cite|improve this question














      Using the Mean Value Theorem, prove that



      $|cos^2(b)-cos^2(a)|gt frac14|b-a| $



      for all $a,b in (fracpi4,fracpi3)$



      So far, I have



      Let $f(x)=cos^2(x)$



      Then $f(x)$ is continuous and differentiable on all $xin mathbbR$ so it is continuous and differentiable on $xin (fracpi4,fracpi3)$.



      Applying the Mean Value Theorem to $f(x)$, we have:



      $fraccos^2(b)-cos^2(a)b-a= -2cos(x)sin(x)$



      $fraccos^2(b)-cos^2(a)b-a= -sin(2x)$



      $|cos^2(b)-cos^2(a)| = -sin(2x)|b-a|$



      Now I do not know how to go from here, as on the interval $(fracpi4,fracpi3)$, $-fracsqrt32lt-sin(2x)lt -1$.



      Or am I making a mistake?





      real-analysis





      share|cite|improve this question













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      edited Aug 20 at 13:41









      Bernard

      111k635103




      111k635103










      asked Aug 20 at 12:55









      user499701

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          You have forgotten the absolute value signs and in the interval $$fracpi4<x<fracpi3$$ is $$|sin(2x)|$$ greater than $$frac14$$
          $$|sin(2x)|$$ has the lowest value $$|sin(frac2pi3)|>frac14$$






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            You have forgotten the absolute value signs and in the interval $$fracpi4<x<fracpi3$$ is $$|sin(2x)|$$ greater than $$frac14$$
            $$|sin(2x)|$$ has the lowest value $$|sin(frac2pi3)|>frac14$$






            share|cite|improve this answer


























              up vote
              2
              down vote













              You have forgotten the absolute value signs and in the interval $$fracpi4<x<fracpi3$$ is $$|sin(2x)|$$ greater than $$frac14$$
              $$|sin(2x)|$$ has the lowest value $$|sin(frac2pi3)|>frac14$$






              share|cite|improve this answer
























                up vote
                2
                down vote










                up vote
                2
                down vote









                You have forgotten the absolute value signs and in the interval $$fracpi4<x<fracpi3$$ is $$|sin(2x)|$$ greater than $$frac14$$
                $$|sin(2x)|$$ has the lowest value $$|sin(frac2pi3)|>frac14$$






                share|cite|improve this answer














                You have forgotten the absolute value signs and in the interval $$fracpi4<x<fracpi3$$ is $$|sin(2x)|$$ greater than $$frac14$$
                $$|sin(2x)|$$ has the lowest value $$|sin(frac2pi3)|>frac14$$







                share|cite|improve this answer














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                edited Aug 20 at 13:13

























                answered Aug 20 at 13:07









                Dr. Sonnhard Graubner

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