Show $sin(pi(1-s)) = sin(pi s)$.












0












$begingroup$


In Stein's "Complex analysis" he uses in the proof of the reflection formula (p. 164 in https://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf )
the following equality:



For $0<s<1$ we have $$frac{pi}{sin pi(1-s)}=frac{pi}{sinpi s}.$$
How can this be true? We have $sin(pi (1-s))=sin(-pi s)=-sin(pi s)$.
If this equation holds for $0<s<1$ then it should on all of $mathbb{R}$. How is this not a contradiction?
Where am I wrong ? Or is it the textbook?






complex-analysis trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you try with $s=frac{1}{2}$, what is happening ? Why ? and with $s=frac{3}{2}$ ?
    $endgroup$
    – Xoff
    Jan 19 '17 at 10:36








  • 2




    $begingroup$
    $sin(pi-alpha)=sinalpha$ is a very basic formula of trigonometry. Apply it for $alpha=pi s$.
    $endgroup$
    – egreg
    Jan 19 '17 at 13:57
















0












$begingroup$


In Stein's "Complex analysis" he uses in the proof of the reflection formula (p. 164 in https://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf )
the following equality:



For $0<s<1$ we have $$frac{pi}{sin pi(1-s)}=frac{pi}{sinpi s}.$$
How can this be true? We have $sin(pi (1-s))=sin(-pi s)=-sin(pi s)$.
If this equation holds for $0<s<1$ then it should on all of $mathbb{R}$. How is this not a contradiction?
Where am I wrong ? Or is it the textbook?






complex-analysis trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you try with $s=frac{1}{2}$, what is happening ? Why ? and with $s=frac{3}{2}$ ?
    $endgroup$
    – Xoff
    Jan 19 '17 at 10:36








  • 2




    $begingroup$
    $sin(pi-alpha)=sinalpha$ is a very basic formula of trigonometry. Apply it for $alpha=pi s$.
    $endgroup$
    – egreg
    Jan 19 '17 at 13:57














0












0








0





$begingroup$


In Stein's "Complex analysis" he uses in the proof of the reflection formula (p. 164 in https://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf )
the following equality:



For $0<s<1$ we have $$frac{pi}{sin pi(1-s)}=frac{pi}{sinpi s}.$$
How can this be true? We have $sin(pi (1-s))=sin(-pi s)=-sin(pi s)$.
If this equation holds for $0<s<1$ then it should on all of $mathbb{R}$. How is this not a contradiction?
Where am I wrong ? Or is it the textbook?






complex-analysis trigonometry







share|cite|improve this question











$endgroup$




In Stein's "Complex analysis" he uses in the proof of the reflection formula (p. 164 in https://www.fing.edu.uy/~cerminar/Complex_Analysis.pdf )
the following equality:



For $0<s<1$ we have $$frac{pi}{sin pi(1-s)}=frac{pi}{sinpi s}.$$
How can this be true? We have $sin(pi (1-s))=sin(-pi s)=-sin(pi s)$.
If this equation holds for $0<s<1$ then it should on all of $mathbb{R}$. How is this not a contradiction?
Where am I wrong ? Or is it the textbook?






complex-analysis trigonometry




complex-analysis trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 17:04









amWhy

1




1










asked Jan 19 '17 at 10:33









RüdigerRüdiger

15819




15819












  • $begingroup$
    If you try with $s=frac{1}{2}$, what is happening ? Why ? and with $s=frac{3}{2}$ ?
    $endgroup$
    – Xoff
    Jan 19 '17 at 10:36








  • 2




    $begingroup$
    $sin(pi-alpha)=sinalpha$ is a very basic formula of trigonometry. Apply it for $alpha=pi s$.
    $endgroup$
    – egreg
    Jan 19 '17 at 13:57


















  • $begingroup$
    If you try with $s=frac{1}{2}$, what is happening ? Why ? and with $s=frac{3}{2}$ ?
    $endgroup$
    – Xoff
    Jan 19 '17 at 10:36








  • 2




    $begingroup$
    $sin(pi-alpha)=sinalpha$ is a very basic formula of trigonometry. Apply it for $alpha=pi s$.
    $endgroup$
    – egreg
    Jan 19 '17 at 13:57
















$begingroup$
If you try with $s=frac{1}{2}$, what is happening ? Why ? and with $s=frac{3}{2}$ ?
$endgroup$
– Xoff
Jan 19 '17 at 10:36






$begingroup$
If you try with $s=frac{1}{2}$, what is happening ? Why ? and with $s=frac{3}{2}$ ?
$endgroup$
– Xoff
Jan 19 '17 at 10:36






2




2




$begingroup$
$sin(pi-alpha)=sinalpha$ is a very basic formula of trigonometry. Apply it for $alpha=pi s$.
$endgroup$
– egreg
Jan 19 '17 at 13:57




$begingroup$
$sin(pi-alpha)=sinalpha$ is a very basic formula of trigonometry. Apply it for $alpha=pi s$.
$endgroup$
– egreg
Jan 19 '17 at 13:57










4 Answers
4






active

oldest

votes


















3












$begingroup$

The textbook is correct and you have a mistake on your analysis. In fact, the equality is rather easy to show:



$sin(pi(1-s)) = sin(pi - pi s)$ and using the trivial fact that $sin(pi - alpha)=sin(alpha)$, you have $sin(pi - pi s) = sin(pi s)$ as argued by the author.



Moreover, the equality holds for any $s$ in the domain of the expression, but beware that any $s in mathbb{Z}$, and in particular $s=0$ and $s=1$, are not in the domain of $frac{pi}{sin(pi s)}$ and so you should not say that the equality holds for any real $s$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Hint: $$sin(pi(1-s))=sin(pi-pi s)=sin(pi)cos(pi s)-sin(pi s)cos(pi)=0 cos(pi s)-sin(pi s)(-1)=sin(pi s)$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
      $endgroup$
      – Rüdiger
      Jan 19 '17 at 12:30










    • $begingroup$
      @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
      $endgroup$
      – MrYouMath
      Jan 20 '17 at 8:45










    • $begingroup$
      you are right! sorry. +1
      $endgroup$
      – Rüdiger
      Jan 20 '17 at 10:42



















    0












    $begingroup$

    $sin(pi - pi s) = sin (pi s) $ because of the fact that sine is positive in second quadrant.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Just look at the sine curve on the domain $[0, pi]$. It has a vertical line of symmetry about $s = frac{pi}{2}$. You have mixed up your trig identities: it is true that cos$pi(1-s) = -$cos$pi s$ for $s in [0, 1]$.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        The textbook is correct and you have a mistake on your analysis. In fact, the equality is rather easy to show:



        $sin(pi(1-s)) = sin(pi - pi s)$ and using the trivial fact that $sin(pi - alpha)=sin(alpha)$, you have $sin(pi - pi s) = sin(pi s)$ as argued by the author.



        Moreover, the equality holds for any $s$ in the domain of the expression, but beware that any $s in mathbb{Z}$, and in particular $s=0$ and $s=1$, are not in the domain of $frac{pi}{sin(pi s)}$ and so you should not say that the equality holds for any real $s$.






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          The textbook is correct and you have a mistake on your analysis. In fact, the equality is rather easy to show:



          $sin(pi(1-s)) = sin(pi - pi s)$ and using the trivial fact that $sin(pi - alpha)=sin(alpha)$, you have $sin(pi - pi s) = sin(pi s)$ as argued by the author.



          Moreover, the equality holds for any $s$ in the domain of the expression, but beware that any $s in mathbb{Z}$, and in particular $s=0$ and $s=1$, are not in the domain of $frac{pi}{sin(pi s)}$ and so you should not say that the equality holds for any real $s$.






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            The textbook is correct and you have a mistake on your analysis. In fact, the equality is rather easy to show:



            $sin(pi(1-s)) = sin(pi - pi s)$ and using the trivial fact that $sin(pi - alpha)=sin(alpha)$, you have $sin(pi - pi s) = sin(pi s)$ as argued by the author.



            Moreover, the equality holds for any $s$ in the domain of the expression, but beware that any $s in mathbb{Z}$, and in particular $s=0$ and $s=1$, are not in the domain of $frac{pi}{sin(pi s)}$ and so you should not say that the equality holds for any real $s$.






            share|cite|improve this answer











            $endgroup$



            The textbook is correct and you have a mistake on your analysis. In fact, the equality is rather easy to show:



            $sin(pi(1-s)) = sin(pi - pi s)$ and using the trivial fact that $sin(pi - alpha)=sin(alpha)$, you have $sin(pi - pi s) = sin(pi s)$ as argued by the author.



            Moreover, the equality holds for any $s$ in the domain of the expression, but beware that any $s in mathbb{Z}$, and in particular $s=0$ and $s=1$, are not in the domain of $frac{pi}{sin(pi s)}$ and so you should not say that the equality holds for any real $s$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Feb 8 '17 at 16:00

























            answered Jan 19 '17 at 13:16









            D...D...

            213113




            213113























                3












                $begingroup$

                Hint: $$sin(pi(1-s))=sin(pi-pi s)=sin(pi)cos(pi s)-sin(pi s)cos(pi)=0 cos(pi s)-sin(pi s)(-1)=sin(pi s)$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
                  $endgroup$
                  – Rüdiger
                  Jan 19 '17 at 12:30










                • $begingroup$
                  @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
                  $endgroup$
                  – MrYouMath
                  Jan 20 '17 at 8:45










                • $begingroup$
                  you are right! sorry. +1
                  $endgroup$
                  – Rüdiger
                  Jan 20 '17 at 10:42
















                3












                $begingroup$

                Hint: $$sin(pi(1-s))=sin(pi-pi s)=sin(pi)cos(pi s)-sin(pi s)cos(pi)=0 cos(pi s)-sin(pi s)(-1)=sin(pi s)$$






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
                  $endgroup$
                  – Rüdiger
                  Jan 19 '17 at 12:30










                • $begingroup$
                  @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
                  $endgroup$
                  – MrYouMath
                  Jan 20 '17 at 8:45










                • $begingroup$
                  you are right! sorry. +1
                  $endgroup$
                  – Rüdiger
                  Jan 20 '17 at 10:42














                3












                3








                3





                $begingroup$

                Hint: $$sin(pi(1-s))=sin(pi-pi s)=sin(pi)cos(pi s)-sin(pi s)cos(pi)=0 cos(pi s)-sin(pi s)(-1)=sin(pi s)$$






                share|cite|improve this answer









                $endgroup$



                Hint: $$sin(pi(1-s))=sin(pi-pi s)=sin(pi)cos(pi s)-sin(pi s)cos(pi)=0 cos(pi s)-sin(pi s)(-1)=sin(pi s)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 '17 at 10:37









                MrYouMathMrYouMath

                13.9k31236




                13.9k31236












                • $begingroup$
                  I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
                  $endgroup$
                  – Rüdiger
                  Jan 19 '17 at 12:30










                • $begingroup$
                  @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
                  $endgroup$
                  – MrYouMath
                  Jan 20 '17 at 8:45










                • $begingroup$
                  you are right! sorry. +1
                  $endgroup$
                  – Rüdiger
                  Jan 20 '17 at 10:42


















                • $begingroup$
                  I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
                  $endgroup$
                  – Rüdiger
                  Jan 19 '17 at 12:30










                • $begingroup$
                  @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
                  $endgroup$
                  – MrYouMath
                  Jan 20 '17 at 8:45










                • $begingroup$
                  you are right! sorry. +1
                  $endgroup$
                  – Rüdiger
                  Jan 20 '17 at 10:42
















                $begingroup$
                I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
                $endgroup$
                – Rüdiger
                Jan 19 '17 at 12:30




                $begingroup$
                I don't think you can argue like this... one needs that $0<s<1$. I don't see where you are using this
                $endgroup$
                – Rüdiger
                Jan 19 '17 at 12:30












                $begingroup$
                @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
                $endgroup$
                – MrYouMath
                Jan 20 '17 at 8:45




                $begingroup$
                @Rüdiger These are standard addition theorems for the sine function, they are true for all values of $s$ (mathworld.wolfram.com/TrigonometricAdditionFormulas.html).
                $endgroup$
                – MrYouMath
                Jan 20 '17 at 8:45












                $begingroup$
                you are right! sorry. +1
                $endgroup$
                – Rüdiger
                Jan 20 '17 at 10:42




                $begingroup$
                you are right! sorry. +1
                $endgroup$
                – Rüdiger
                Jan 20 '17 at 10:42











                0












                $begingroup$

                $sin(pi - pi s) = sin (pi s) $ because of the fact that sine is positive in second quadrant.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  $sin(pi - pi s) = sin (pi s) $ because of the fact that sine is positive in second quadrant.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    $sin(pi - pi s) = sin (pi s) $ because of the fact that sine is positive in second quadrant.






                    share|cite|improve this answer









                    $endgroup$



                    $sin(pi - pi s) = sin (pi s) $ because of the fact that sine is positive in second quadrant.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 19 '17 at 10:37









                    FawadFawad

                    1,40111234




                    1,40111234























                        0












                        $begingroup$

                        Just look at the sine curve on the domain $[0, pi]$. It has a vertical line of symmetry about $s = frac{pi}{2}$. You have mixed up your trig identities: it is true that cos$pi(1-s) = -$cos$pi s$ for $s in [0, 1]$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Just look at the sine curve on the domain $[0, pi]$. It has a vertical line of symmetry about $s = frac{pi}{2}$. You have mixed up your trig identities: it is true that cos$pi(1-s) = -$cos$pi s$ for $s in [0, 1]$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Just look at the sine curve on the domain $[0, pi]$. It has a vertical line of symmetry about $s = frac{pi}{2}$. You have mixed up your trig identities: it is true that cos$pi(1-s) = -$cos$pi s$ for $s in [0, 1]$.






                            share|cite|improve this answer









                            $endgroup$



                            Just look at the sine curve on the domain $[0, pi]$. It has a vertical line of symmetry about $s = frac{pi}{2}$. You have mixed up your trig identities: it is true that cos$pi(1-s) = -$cos$pi s$ for $s in [0, 1]$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 19 '17 at 12:59









                            Vik78Vik78

                            2,164616




                            2,164616






























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