Prove the definition of the arcsin(s).












2












$begingroup$


I am given $arcsin: S rightarrow (-pi/2,pi/2) $ is the inverse function of sin(t) (restricted to [$-pi/2,pi/2$]). I'm trying to prove that $arcsin(s)$= $int_{0}^{s}1/sqrt{1-x^2}$ .



My initial thoughts for an attempt of a proof:



By definition, we know:



$sin(arcsin(x)) = x$



Thus by the chain rule:



$(sin(arcsin(x)) = x)' $



$ cos(arcsin(x)) times d (arcsin(x))/dx = 1$



Also by definition, we know: $sin(x)^2 + cos(x)^2 = 1$



It has been given that $cos(x) > 0 $ over the interval $(-pi/2,pi/2)$, which implies



$sin(arcsin(x))^2 + cos(arcsin(x))^2= 1$.



Therefore, $cos(arcsin(x)) = sqrt{(1-x^2)}$



So, $sqrt{(1-x^2)}times d (arcsin(x)) /dx = 1$ and



$d (arcsin(x)) /dx = 1/sqrt{(1-x^2)}$



From there I think we will use parts of the MVT or the fundamental theorem of calculus to prove the result $arcsin(s)=int_{0}^{s}1/sqrt{1-x^2}$, but I'm having trouble with the formalities.






real-analysis calculus trigonometry proof-writing chain-rule







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you just integrate both side by x from 0 to s?
    $endgroup$
    – KYHSGeekCode
    Dec 9 '18 at 23:46










  • $begingroup$
    The fundamental theorem seems to be the way to go. At least one answer below uses it implicitly.
    $endgroup$
    – David K
    Dec 10 '18 at 0:41
















2












$begingroup$


I am given $arcsin: S rightarrow (-pi/2,pi/2) $ is the inverse function of sin(t) (restricted to [$-pi/2,pi/2$]). I'm trying to prove that $arcsin(s)$= $int_{0}^{s}1/sqrt{1-x^2}$ .



My initial thoughts for an attempt of a proof:



By definition, we know:



$sin(arcsin(x)) = x$



Thus by the chain rule:



$(sin(arcsin(x)) = x)' $



$ cos(arcsin(x)) times d (arcsin(x))/dx = 1$



Also by definition, we know: $sin(x)^2 + cos(x)^2 = 1$



It has been given that $cos(x) > 0 $ over the interval $(-pi/2,pi/2)$, which implies



$sin(arcsin(x))^2 + cos(arcsin(x))^2= 1$.



Therefore, $cos(arcsin(x)) = sqrt{(1-x^2)}$



So, $sqrt{(1-x^2)}times d (arcsin(x)) /dx = 1$ and



$d (arcsin(x)) /dx = 1/sqrt{(1-x^2)}$



From there I think we will use parts of the MVT or the fundamental theorem of calculus to prove the result $arcsin(s)=int_{0}^{s}1/sqrt{1-x^2}$, but I'm having trouble with the formalities.






real-analysis calculus trigonometry proof-writing chain-rule







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you just integrate both side by x from 0 to s?
    $endgroup$
    – KYHSGeekCode
    Dec 9 '18 at 23:46










  • $begingroup$
    The fundamental theorem seems to be the way to go. At least one answer below uses it implicitly.
    $endgroup$
    – David K
    Dec 10 '18 at 0:41














2












2








2





$begingroup$


I am given $arcsin: S rightarrow (-pi/2,pi/2) $ is the inverse function of sin(t) (restricted to [$-pi/2,pi/2$]). I'm trying to prove that $arcsin(s)$= $int_{0}^{s}1/sqrt{1-x^2}$ .



My initial thoughts for an attempt of a proof:



By definition, we know:



$sin(arcsin(x)) = x$



Thus by the chain rule:



$(sin(arcsin(x)) = x)' $



$ cos(arcsin(x)) times d (arcsin(x))/dx = 1$



Also by definition, we know: $sin(x)^2 + cos(x)^2 = 1$



It has been given that $cos(x) > 0 $ over the interval $(-pi/2,pi/2)$, which implies



$sin(arcsin(x))^2 + cos(arcsin(x))^2= 1$.



Therefore, $cos(arcsin(x)) = sqrt{(1-x^2)}$



So, $sqrt{(1-x^2)}times d (arcsin(x)) /dx = 1$ and



$d (arcsin(x)) /dx = 1/sqrt{(1-x^2)}$



From there I think we will use parts of the MVT or the fundamental theorem of calculus to prove the result $arcsin(s)=int_{0}^{s}1/sqrt{1-x^2}$, but I'm having trouble with the formalities.






real-analysis calculus trigonometry proof-writing chain-rule







share|cite|improve this question











$endgroup$




I am given $arcsin: S rightarrow (-pi/2,pi/2) $ is the inverse function of sin(t) (restricted to [$-pi/2,pi/2$]). I'm trying to prove that $arcsin(s)$= $int_{0}^{s}1/sqrt{1-x^2}$ .



My initial thoughts for an attempt of a proof:



By definition, we know:



$sin(arcsin(x)) = x$



Thus by the chain rule:



$(sin(arcsin(x)) = x)' $



$ cos(arcsin(x)) times d (arcsin(x))/dx = 1$



Also by definition, we know: $sin(x)^2 + cos(x)^2 = 1$



It has been given that $cos(x) > 0 $ over the interval $(-pi/2,pi/2)$, which implies



$sin(arcsin(x))^2 + cos(arcsin(x))^2= 1$.



Therefore, $cos(arcsin(x)) = sqrt{(1-x^2)}$



So, $sqrt{(1-x^2)}times d (arcsin(x)) /dx = 1$ and



$d (arcsin(x)) /dx = 1/sqrt{(1-x^2)}$



From there I think we will use parts of the MVT or the fundamental theorem of calculus to prove the result $arcsin(s)=int_{0}^{s}1/sqrt{1-x^2}$, but I'm having trouble with the formalities.






real-analysis calculus trigonometry proof-writing chain-rule




real-analysis calculus trigonometry proof-writing chain-rule






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 0:32









Bernard

123k741117




123k741117










asked Dec 9 '18 at 23:35







user624612



















  • $begingroup$
    Why don't you just integrate both side by x from 0 to s?
    $endgroup$
    – KYHSGeekCode
    Dec 9 '18 at 23:46










  • $begingroup$
    The fundamental theorem seems to be the way to go. At least one answer below uses it implicitly.
    $endgroup$
    – David K
    Dec 10 '18 at 0:41


















  • $begingroup$
    Why don't you just integrate both side by x from 0 to s?
    $endgroup$
    – KYHSGeekCode
    Dec 9 '18 at 23:46










  • $begingroup$
    The fundamental theorem seems to be the way to go. At least one answer below uses it implicitly.
    $endgroup$
    – David K
    Dec 10 '18 at 0:41
















$begingroup$
Why don't you just integrate both side by x from 0 to s?
$endgroup$
– KYHSGeekCode
Dec 9 '18 at 23:46




$begingroup$
Why don't you just integrate both side by x from 0 to s?
$endgroup$
– KYHSGeekCode
Dec 9 '18 at 23:46












$begingroup$
The fundamental theorem seems to be the way to go. At least one answer below uses it implicitly.
$endgroup$
– David K
Dec 10 '18 at 0:41




$begingroup$
The fundamental theorem seems to be the way to go. At least one answer below uses it implicitly.
$endgroup$
– David K
Dec 10 '18 at 0:41










2 Answers
2






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oldest

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1












$begingroup$

You can use the fact that$$arcsin'(x)=frac1{sin'(arcsin x)}=frac1{cos(arcsin x)}=frac1{sqrt{1-x^2}}.$$So, and since $arcsin(0)=0$, you get that$$arcsin s=int_0^sfrac{mathrm dx}{sqrt{1-x^2}}.$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Problem: $$int_0^sfrac 1{sqrt{1-x^2}}dx$$



    Let $$x=sin theta$$ then we get:
    $$frac{dx}{dtheta}=cos theta$$
    By derivating both side by $x$.



    $$therefore int_0^sfrac 1{sqrt{1-x^2}}dx$$
    $$=int_0^{arcsin s}frac {costheta}{sqrt{1-{sin^2 theta}}}dtheta$$
    $$=int_0^{arcsin s} 1 dtheta$$
    $$=left[ theta right]^{arcsin s}_{arcsin 0} dtheta$$
    $$=arcsin s$$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

      oldest

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      1












      $begingroup$

      You can use the fact that$$arcsin'(x)=frac1{sin'(arcsin x)}=frac1{cos(arcsin x)}=frac1{sqrt{1-x^2}}.$$So, and since $arcsin(0)=0$, you get that$$arcsin s=int_0^sfrac{mathrm dx}{sqrt{1-x^2}}.$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        You can use the fact that$$arcsin'(x)=frac1{sin'(arcsin x)}=frac1{cos(arcsin x)}=frac1{sqrt{1-x^2}}.$$So, and since $arcsin(0)=0$, you get that$$arcsin s=int_0^sfrac{mathrm dx}{sqrt{1-x^2}}.$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          You can use the fact that$$arcsin'(x)=frac1{sin'(arcsin x)}=frac1{cos(arcsin x)}=frac1{sqrt{1-x^2}}.$$So, and since $arcsin(0)=0$, you get that$$arcsin s=int_0^sfrac{mathrm dx}{sqrt{1-x^2}}.$$






          share|cite|improve this answer









          $endgroup$



          You can use the fact that$$arcsin'(x)=frac1{sin'(arcsin x)}=frac1{cos(arcsin x)}=frac1{sqrt{1-x^2}}.$$So, and since $arcsin(0)=0$, you get that$$arcsin s=int_0^sfrac{mathrm dx}{sqrt{1-x^2}}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 23:43









          José Carlos SantosJosé Carlos Santos

          169k23132237




          169k23132237























              0












              $begingroup$

              Problem: $$int_0^sfrac 1{sqrt{1-x^2}}dx$$



              Let $$x=sin theta$$ then we get:
              $$frac{dx}{dtheta}=cos theta$$
              By derivating both side by $x$.



              $$therefore int_0^sfrac 1{sqrt{1-x^2}}dx$$
              $$=int_0^{arcsin s}frac {costheta}{sqrt{1-{sin^2 theta}}}dtheta$$
              $$=int_0^{arcsin s} 1 dtheta$$
              $$=left[ theta right]^{arcsin s}_{arcsin 0} dtheta$$
              $$=arcsin s$$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                Problem: $$int_0^sfrac 1{sqrt{1-x^2}}dx$$



                Let $$x=sin theta$$ then we get:
                $$frac{dx}{dtheta}=cos theta$$
                By derivating both side by $x$.



                $$therefore int_0^sfrac 1{sqrt{1-x^2}}dx$$
                $$=int_0^{arcsin s}frac {costheta}{sqrt{1-{sin^2 theta}}}dtheta$$
                $$=int_0^{arcsin s} 1 dtheta$$
                $$=left[ theta right]^{arcsin s}_{arcsin 0} dtheta$$
                $$=arcsin s$$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Problem: $$int_0^sfrac 1{sqrt{1-x^2}}dx$$



                  Let $$x=sin theta$$ then we get:
                  $$frac{dx}{dtheta}=cos theta$$
                  By derivating both side by $x$.



                  $$therefore int_0^sfrac 1{sqrt{1-x^2}}dx$$
                  $$=int_0^{arcsin s}frac {costheta}{sqrt{1-{sin^2 theta}}}dtheta$$
                  $$=int_0^{arcsin s} 1 dtheta$$
                  $$=left[ theta right]^{arcsin s}_{arcsin 0} dtheta$$
                  $$=arcsin s$$.






                  share|cite|improve this answer











                  $endgroup$



                  Problem: $$int_0^sfrac 1{sqrt{1-x^2}}dx$$



                  Let $$x=sin theta$$ then we get:
                  $$frac{dx}{dtheta}=cos theta$$
                  By derivating both side by $x$.



                  $$therefore int_0^sfrac 1{sqrt{1-x^2}}dx$$
                  $$=int_0^{arcsin s}frac {costheta}{sqrt{1-{sin^2 theta}}}dtheta$$
                  $$=int_0^{arcsin s} 1 dtheta$$
                  $$=left[ theta right]^{arcsin s}_{arcsin 0} dtheta$$
                  $$=arcsin s$$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 9 '18 at 23:48

























                  answered Dec 9 '18 at 23:42









                  KYHSGeekCodeKYHSGeekCode

                  334112




                  334112






























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