Prove $sin^{-1}(1)geq int_0^b1/sqrt{1-x^2}dx +(1-b)pi/2$ for $b in [0,1)$












3












$begingroup$


I'm trying to prove the following inequality:



$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$



for every $b in [0,1)$.



I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).



My Attempt



I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.



This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.



I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.






real-analysis integration trigonometry inequality proof-writing







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  • 3




    $begingroup$
    Small thing: it should be $le$ not $<$ because of $b=0$.
    $endgroup$
    – user608030
    Dec 11 '18 at 17:51










  • $begingroup$
    It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
    $endgroup$
    – Sameer Baheti
    Dec 11 '18 at 18:07


















3












$begingroup$


I'm trying to prove the following inequality:



$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$



for every $b in [0,1)$.



I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).



My Attempt



I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.



This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.



I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.






real-analysis integration trigonometry inequality proof-writing







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Small thing: it should be $le$ not $<$ because of $b=0$.
    $endgroup$
    – user608030
    Dec 11 '18 at 17:51










  • $begingroup$
    It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
    $endgroup$
    – Sameer Baheti
    Dec 11 '18 at 18:07
















3












3








3





$begingroup$


I'm trying to prove the following inequality:



$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$



for every $b in [0,1)$.



I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).



My Attempt



I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.



This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.



I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.






real-analysis integration trigonometry inequality proof-writing







share|cite|improve this question











$endgroup$




I'm trying to prove the following inequality:



$$sin^{-1}(1)geqint_0^b1/sqrt{1-x^2},dx +(1-b)pi/2$$



for every $b in [0,1)$.



I'm given $sin^{-1}(1) = pi/2$ and $sin^{-1}(x)$ is strictly increasing. We also know $sin^{-1}(x)$ is the inverse of the strictly increasing function $sin(x)$ (when $xin [-pi/2, pi/2] $).



My Attempt



I can prove using integration and the FTC that
$int_0^b1/sqrt{1-x^2},dx = sin^{-1}(b)$.



This information simplifies the inequality to $0 geq sin^{-1}(b) - btimes pi/2$.



I'm having trouble showing that $ sin^{-1}(b) leq btimes pi/2$ given that everything above is true.






real-analysis integration trigonometry inequality proof-writing




real-analysis integration trigonometry inequality proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 7:44









Sameer Baheti

5618




5618










asked Dec 11 '18 at 17:39







user624612















  • 3




    $begingroup$
    Small thing: it should be $le$ not $<$ because of $b=0$.
    $endgroup$
    – user608030
    Dec 11 '18 at 17:51










  • $begingroup$
    It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
    $endgroup$
    – Sameer Baheti
    Dec 11 '18 at 18:07
















  • 3




    $begingroup$
    Small thing: it should be $le$ not $<$ because of $b=0$.
    $endgroup$
    – user608030
    Dec 11 '18 at 17:51










  • $begingroup$
    It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
    $endgroup$
    – Sameer Baheti
    Dec 11 '18 at 18:07










3




3




$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– user608030
Dec 11 '18 at 17:51




$begingroup$
Small thing: it should be $le$ not $<$ because of $b=0$.
$endgroup$
– user608030
Dec 11 '18 at 17:51












$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07






$begingroup$
It should be $geq$ in the title and $arcsin(b) leq btimes pi/2$ in the body if $b in [0,frac{pi}2)$
$endgroup$
– Sameer Baheti
Dec 11 '18 at 18:07












5 Answers
5






active

oldest

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1












$begingroup$

The inequality you want to prove is
$$
arcsin1-arcsin b<frac{pi}{2}(1-b).
$$
This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
$$
arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
$$
If
$$
frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
$$
then
$$
b<xi<sqrt{1-frac{4}{pi^2}}.
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      The inequality you seek to prove is not true. Put $b=1/4$.



      However, $sin^{-1}ble bpi/2, forall bin[0,1)$



      Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$



      $f'(x)=pi/2-frac1{sqrt{1-x^2}}$



      $f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$



      This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
      This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$






      share|cite|improve this answer











      $endgroup$





















        0












        $begingroup$

        If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$  This implies



        $$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$



        for $bin (0,1).$ Apply this with $f(b)=arcsin b.$






        share|cite|improve this answer









        $endgroup$





















          0












          $begingroup$

          Your inequality is reversed. The correct inequality is
          $$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
          for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.



          As you did, we can show that the inequality above is equivalent to
          $$arcsin(b)leq frac{pi}{2},b$$
          for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
          $$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
          by Jensen's Inequality. This shows that
          $$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
          Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.






          share|cite|improve this answer









          $endgroup$














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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The inequality you want to prove is
            $$
            arcsin1-arcsin b<frac{pi}{2}(1-b).
            $$
            This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
            $$
            arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
            $$
            If
            $$
            frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
            $$
            then
            $$
            b<xi<sqrt{1-frac{4}{pi^2}}.
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The inequality you want to prove is
              $$
              arcsin1-arcsin b<frac{pi}{2}(1-b).
              $$
              This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
              $$
              arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
              $$
              If
              $$
              frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
              $$
              then
              $$
              b<xi<sqrt{1-frac{4}{pi^2}}.
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The inequality you want to prove is
                $$
                arcsin1-arcsin b<frac{pi}{2}(1-b).
                $$
                This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
                $$
                arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
                $$
                If
                $$
                frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
                $$
                then
                $$
                b<xi<sqrt{1-frac{4}{pi^2}}.
                $$






                share|cite|improve this answer









                $endgroup$



                The inequality you want to prove is
                $$
                arcsin1-arcsin b<frac{pi}{2}(1-b).
                $$
                This inequality is false for $b$ close to $1$. Indeed, by the Mean Value Theorem
                $$
                arcsin1-arcsin b=frac{1-b}{sqrt{1-xi^2}},quad b<xi<1.
                $$
                If
                $$
                frac{1-b}{sqrt{1-xi^2}}<frac{pi}{2}(1-b),
                $$
                then
                $$
                b<xi<sqrt{1-frac{4}{pi^2}}.
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 11 '18 at 18:21









                Julián AguirreJulián Aguirre

                69.5k24297




                69.5k24297























                    0












                    $begingroup$

                    I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.






                    share|cite|improve this answer











                    $endgroup$


















                      0












                      $begingroup$

                      I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.






                      share|cite|improve this answer











                      $endgroup$
















                        0












                        0








                        0





                        $begingroup$

                        I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.






                        share|cite|improve this answer











                        $endgroup$



                        I think $$sin^{-1}(b) leq bfrac{pi}2$$ for every $b in big[0,1)$ since the graph of $y=b$ when scaled upto $y=frac{pi}2b $ will get above that of $sin^{-1}(b)$ for $binbig(0,1)$ and equal for $b=0$.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 11 '18 at 18:12

























                        answered Dec 11 '18 at 18:03









                        Sameer BahetiSameer Baheti

                        5618




                        5618























                            0












                            $begingroup$

                            The inequality you seek to prove is not true. Put $b=1/4$.



                            However, $sin^{-1}ble bpi/2, forall bin[0,1)$



                            Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$



                            $f'(x)=pi/2-frac1{sqrt{1-x^2}}$



                            $f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$



                            This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
                            This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$






                            share|cite|improve this answer











                            $endgroup$


















                              0












                              $begingroup$

                              The inequality you seek to prove is not true. Put $b=1/4$.



                              However, $sin^{-1}ble bpi/2, forall bin[0,1)$



                              Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$



                              $f'(x)=pi/2-frac1{sqrt{1-x^2}}$



                              $f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$



                              This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
                              This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$






                              share|cite|improve this answer











                              $endgroup$
















                                0












                                0








                                0





                                $begingroup$

                                The inequality you seek to prove is not true. Put $b=1/4$.



                                However, $sin^{-1}ble bpi/2, forall bin[0,1)$



                                Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$



                                $f'(x)=pi/2-frac1{sqrt{1-x^2}}$



                                $f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$



                                This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
                                This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$






                                share|cite|improve this answer











                                $endgroup$



                                The inequality you seek to prove is not true. Put $b=1/4$.



                                However, $sin^{-1}ble bpi/2, forall bin[0,1)$



                                Let $f(x)=xpi/2-sin^{-1}x, f(0)=f(1)=0$



                                $f'(x)=pi/2-frac1{sqrt{1-x^2}}$



                                $f'(x)>0, forall xinBig[0,sqrt{1-frac4{pi^2}}Big)$ and $f'(x)<0, forall xinBig(sqrt{1-frac4{pi^2}},1Big)$



                                This means that $f(x)$ initally increases from $f(0)$ to a maxima and then decreases to $f(1)$ in $[0,1]$. So $f(x)$ stays above $min{f(0),f(1)}=0$.
                                This means $f(x)ge0 forall xin[0,1)implies f(b)ge0$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 11 '18 at 18:29

























                                answered Dec 11 '18 at 17:57









                                Shubham JohriShubham Johri

                                5,475818




                                5,475818























                                    0












                                    $begingroup$

                                    If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$  This implies



                                    $$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$



                                    for $bin (0,1).$ Apply this with $f(b)=arcsin b.$






                                    share|cite|improve this answer









                                    $endgroup$


















                                      0












                                      $begingroup$

                                      If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$  This implies



                                      $$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$



                                      for $bin (0,1).$ Apply this with $f(b)=arcsin b.$






                                      share|cite|improve this answer









                                      $endgroup$
















                                        0












                                        0








                                        0





                                        $begingroup$

                                        If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$  This implies



                                        $$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$



                                        for $bin (0,1).$ Apply this with $f(b)=arcsin b.$






                                        share|cite|improve this answer









                                        $endgroup$



                                        If $f$ is strictly convex on $[0,1],$ then $(b,f(b))$ lies below the line through $(0,f(0))$ and $(1,f(1))$ for $bin (0,1).$  This implies



                                        $$frac{f(1)-f(b)}{1-b} > frac{f(1)-f(0)}{1}$$



                                        for $bin (0,1).$ Apply this with $f(b)=arcsin b.$







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Dec 11 '18 at 19:29









                                        zhw.zhw.

                                        74.7k43275




                                        74.7k43275























                                            0












                                            $begingroup$

                                            Your inequality is reversed. The correct inequality is
                                            $$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
                                            for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.



                                            As you did, we can show that the inequality above is equivalent to
                                            $$arcsin(b)leq frac{pi}{2},b$$
                                            for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
                                            $$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
                                            by Jensen's Inequality. This shows that
                                            $$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
                                            Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              Your inequality is reversed. The correct inequality is
                                              $$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
                                              for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.



                                              As you did, we can show that the inequality above is equivalent to
                                              $$arcsin(b)leq frac{pi}{2},b$$
                                              for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
                                              $$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
                                              by Jensen's Inequality. This shows that
                                              $$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
                                              Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                Your inequality is reversed. The correct inequality is
                                                $$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
                                                for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.



                                                As you did, we can show that the inequality above is equivalent to
                                                $$arcsin(b)leq frac{pi}{2},b$$
                                                for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
                                                $$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
                                                by Jensen's Inequality. This shows that
                                                $$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
                                                Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.






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                                                $endgroup$



                                                Your inequality is reversed. The correct inequality is
                                                $$arcsin(1)geq int_0^b,frac{1}{sqrt{1-x^2}},text{d}x+frac{pi}{2},(1-b)$$
                                                for all $bin[0,1]$. The equality holds if and only if $b=0$ or $b=1$.



                                                As you did, we can show that the inequality above is equivalent to
                                                $$arcsin(b)leq frac{pi}{2},b$$
                                                for all $bin[0,1]$. Now, the function $f:=arcsin$ is convex on $[0,1]$. Therefore, for each $bin [0,1]$,
                                                $$f(b)=fbig((1-b)cdot 0+bcdot 1big)leq (1-b)cdot f(0)+bcdot f(1)$$
                                                by Jensen's Inequality. This shows that
                                                $$arcsin(b)leq (1-b)cdot 0+bcdotfrac{pi}{2}=frac{pi}{2},btext{ for all }bin[0,1],.$$
                                                Since $arcsin$ is strictly convex, the only equality cases are $b=0$ and $b=1$.







                                                share|cite|improve this answer












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                                                answered Dec 11 '18 at 23:26









                                                BatominovskiBatominovski

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