Prove that $arcsin z = frac{pi}{2} - arccos z$












1












$begingroup$


I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis







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  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46
















1












$begingroup$


I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46














1












1








1


1



$begingroup$


I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis







share|cite|improve this question











$endgroup$




I have $arccos (z) = -iln (z + sqrt{z^2-1})$ and $arcsin (z)=-i ln(iz +sqrt{1-z^2}).$



Now I must prove, that $arcsin (z) = frac{pi}{2} - arccos (z)$.



I get:



$$arcsin (z)=-ilnleft(iz+sqrt{1-z^2}right)=-ilnleft(iz+sqrt{-1(z^2-1)}right)=-ilnleft(iz+sqrt{i^{2}(z^2-1)}right)=$$
$$=-lnleft(iz+isqrt{z^2-1}right)=-ilnleft(i(z+sqrt{z^2-1})right)=-iln -ilnleft(z+sqrt{z^2-1}right)=$$
$$=-ileft(ln|i|+ifrac{pi}{2}right)-ilnleft(z+sqrt{z^2-1}right)=frac{pi}{2}+arccos (z).$$



I don't know where I make mistake, because I have "+" not "-".






complex-analysis




complex-analysis






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share|cite|improve this question













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edited May 7 '14 at 14:23









Samrat Mukhopadhyay

13.7k2047




13.7k2047










asked May 7 '14 at 14:11









user119543user119543

115




115












  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46


















  • $begingroup$
    One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
    $endgroup$
    – gammatester
    May 7 '14 at 14:19












  • $begingroup$
    so what I do now
    $endgroup$
    – user119543
    May 7 '14 at 14:26










  • $begingroup$
    Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
    $endgroup$
    – Daniel Fischer♦
    May 7 '14 at 14:27










  • $begingroup$
    but it is multiplicative funkction ... i would do made in this way ..
    $endgroup$
    – user119543
    May 7 '14 at 14:29










  • $begingroup$
    someone can help me?
    $endgroup$
    – user119543
    May 7 '14 at 14:46
















$begingroup$
One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
$endgroup$
– gammatester
May 7 '14 at 14:19






$begingroup$
One hint: $ln(sqrt{i^2})$ is not always equal to $ln(i)$
$endgroup$
– gammatester
May 7 '14 at 14:19














$begingroup$
so what I do now
$endgroup$
– user119543
May 7 '14 at 14:26




$begingroup$
so what I do now
$endgroup$
– user119543
May 7 '14 at 14:26












$begingroup$
Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
$endgroup$
– Daniel Fischer♦
May 7 '14 at 14:27




$begingroup$
Easiest way would be using $sin left(frac{pi}{2}-wright) = cos w$.
$endgroup$
– Daniel Fischer♦
May 7 '14 at 14:27












$begingroup$
but it is multiplicative funkction ... i would do made in this way ..
$endgroup$
– user119543
May 7 '14 at 14:29




$begingroup$
but it is multiplicative funkction ... i would do made in this way ..
$endgroup$
– user119543
May 7 '14 at 14:29












$begingroup$
someone can help me?
$endgroup$
– user119543
May 7 '14 at 14:46




$begingroup$
someone can help me?
$endgroup$
– user119543
May 7 '14 at 14:46










3 Answers
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0












$begingroup$

Suppose the statement is true, namely,
begin{align}
sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
end{align}
By taking the sine of both sides leads to
begin{align}
sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
z &= z.
end{align}
This shows the identity is true.






share|cite|improve this answer









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    0












    $begingroup$

    lets start by defining $w to be $sin(w) = z
    now, we'll use the complex identity:



    sin(w) = z = (exp(iw)-exp(-iw))/2i


    define t = exp(iw) therefor, by applying this to the previous equation:



    z = (t - 1/t)/2i


    multiply by 2i and pass all args to the right side to get:



    t^2 - (2iz) - 1 = 0


    a standard quadratic equation:



    t = ((2iz)+sqrt((2i*z)^2 + 4))/2
    
t = i*z + sqrt(1 - z^2)


    but we defined t = exp(i*w) therfor:



    exp(iw) = i*z + sqrt(1 - z^2)


    take the log to both sides and get that:



    iw = log(i*z + sqrt(1 - z^2))
    divide by i:



    w = (1/i) * log(i*z + sqrt(1 - z^2))


    but as we first defined: sin(w) = z thenw = arcsin(z)
    in total:



    arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


    that proves your claim is right.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Please do not bump old posts unless absolutely necessary.
      $endgroup$
      – Cyclohexanol.
      Dec 3 '14 at 23:14



















    0












    $begingroup$

    My usual naive approach.



    From
    $arccos (z) = -iln (z + sqrt{z^2-1})
    $
    and
    $arcsin (z)=-i ln(iz +sqrt{1-z^2})
    $
    we get



    $begin{array}\
    arccos (z)+arcsin (z)
    &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
    &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
    &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
    &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
    &=-i(ln (i))\
    &=-i(dfrac{ipi}{2})\
    &=dfrac{pi}{2}\
    end{array}
    $



    Throw in a $2kpi i$
    if you want.
    That will change the answer to
    $dfrac{pi}{2}+2kpi$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Suppose the statement is true, namely,
      begin{align}
      sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
      end{align}
      By taking the sine of both sides leads to
      begin{align}
      sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
      z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
      z &= z.
      end{align}
      This shows the identity is true.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Suppose the statement is true, namely,
        begin{align}
        sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
        end{align}
        By taking the sine of both sides leads to
        begin{align}
        sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
        z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
        z &= z.
        end{align}
        This shows the identity is true.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Suppose the statement is true, namely,
          begin{align}
          sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
          end{align}
          By taking the sine of both sides leads to
          begin{align}
          sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
          z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
          z &= z.
          end{align}
          This shows the identity is true.






          share|cite|improve this answer









          $endgroup$



          Suppose the statement is true, namely,
          begin{align}
          sin^{-1}(z) = frac{pi}{2} - cos^{-1}(z).
          end{align}
          By taking the sine of both sides leads to
          begin{align}
          sin(sin^{-1}(z)) &= sin(frac{pi}{2} - cos^{-1}(z)) \
          z &= sin(pi/2) cos(cos^{-1}(z)) - cos(pi/2) sin(cos^{-1}(z)) \
          z &= z.
          end{align}
          This shows the identity is true.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 7 '14 at 17:39









          LeucippusLeucippus

          19.6k102871




          19.6k102871























              0












              $begingroup$

              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14
















              0












              $begingroup$

              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14














              0












              0








              0





              $begingroup$

              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.






              share|cite|improve this answer









              $endgroup$



              lets start by defining $w to be $sin(w) = z
              now, we'll use the complex identity:



              sin(w) = z = (exp(iw)-exp(-iw))/2i


              define t = exp(iw) therefor, by applying this to the previous equation:



              z = (t - 1/t)/2i


              multiply by 2i and pass all args to the right side to get:



              t^2 - (2iz) - 1 = 0


              a standard quadratic equation:



              t = ((2iz)+sqrt((2i*z)^2 + 4))/2
              
t = i*z + sqrt(1 - z^2)


              but we defined t = exp(i*w) therfor:



              exp(iw) = i*z + sqrt(1 - z^2)


              take the log to both sides and get that:



              iw = log(i*z + sqrt(1 - z^2))
              divide by i:



              w = (1/i) * log(i*z + sqrt(1 - z^2))


              but as we first defined: sin(w) = z thenw = arcsin(z)
              in total:



              arcsin(z) = (1/i) * log(i*z + sqrt(1 - z^2))


              that proves your claim is right.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 3 '14 at 23:09









              ThunderWiringThunderWiring

              1215




              1215












              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14


















              • $begingroup$
                Please do not bump old posts unless absolutely necessary.
                $endgroup$
                – Cyclohexanol.
                Dec 3 '14 at 23:14
















              $begingroup$
              Please do not bump old posts unless absolutely necessary.
              $endgroup$
              – Cyclohexanol.
              Dec 3 '14 at 23:14




              $begingroup$
              Please do not bump old posts unless absolutely necessary.
              $endgroup$
              – Cyclohexanol.
              Dec 3 '14 at 23:14











              0












              $begingroup$

              My usual naive approach.



              From
              $arccos (z) = -iln (z + sqrt{z^2-1})
              $
              and
              $arcsin (z)=-i ln(iz +sqrt{1-z^2})
              $
              we get



              $begin{array}\
              arccos (z)+arcsin (z)
              &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
              &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
              &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
              &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
              &=-i(ln (i))\
              &=-i(dfrac{ipi}{2})\
              &=dfrac{pi}{2}\
              end{array}
              $



              Throw in a $2kpi i$
              if you want.
              That will change the answer to
              $dfrac{pi}{2}+2kpi$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                My usual naive approach.



                From
                $arccos (z) = -iln (z + sqrt{z^2-1})
                $
                and
                $arcsin (z)=-i ln(iz +sqrt{1-z^2})
                $
                we get



                $begin{array}\
                arccos (z)+arcsin (z)
                &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
                &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
                &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
                &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
                &=-i(ln (i))\
                &=-i(dfrac{ipi}{2})\
                &=dfrac{pi}{2}\
                end{array}
                $



                Throw in a $2kpi i$
                if you want.
                That will change the answer to
                $dfrac{pi}{2}+2kpi$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  My usual naive approach.



                  From
                  $arccos (z) = -iln (z + sqrt{z^2-1})
                  $
                  and
                  $arcsin (z)=-i ln(iz +sqrt{1-z^2})
                  $
                  we get



                  $begin{array}\
                  arccos (z)+arcsin (z)
                  &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
                  &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
                  &=-i(ln (i))\
                  &=-i(dfrac{ipi}{2})\
                  &=dfrac{pi}{2}\
                  end{array}
                  $



                  Throw in a $2kpi i$
                  if you want.
                  That will change the answer to
                  $dfrac{pi}{2}+2kpi$.






                  share|cite|improve this answer









                  $endgroup$



                  My usual naive approach.



                  From
                  $arccos (z) = -iln (z + sqrt{z^2-1})
                  $
                  and
                  $arcsin (z)=-i ln(iz +sqrt{1-z^2})
                  $
                  we get



                  $begin{array}\
                  arccos (z)+arcsin (z)
                  &=-i(ln (z + sqrt{z^2-1})+ln(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + sqrt{z^2-1})(iz +sqrt{1-z^2}))\
                  &=-i(ln ((z + isqrt{1-z^2})(iz +sqrt{1-z^2}))\
                  &=-i(ln (iz^2+zsqrt{1-z^2}(-1+1)+i(1-z^2))\
                  &=-i(ln (i))\
                  &=-i(dfrac{ipi}{2})\
                  &=dfrac{pi}{2}\
                  end{array}
                  $



                  Throw in a $2kpi i$
                  if you want.
                  That will change the answer to
                  $dfrac{pi}{2}+2kpi$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 21:49









                  marty cohenmarty cohen

                  73.2k549128




                  73.2k549128






























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