Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$?












3












$begingroup$


Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.



It was a multiple choice questions with



a) $4sinh(3) sinh(1)$



b) $4sinh(1)sinh(3)$



I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .



Am I correct both option is correct ?
If wrong please explain why ?






complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You're correct. Multiplication is commutative. The question was badly designed.
    $endgroup$
    – Michael Lugo
    Jan 3 at 15:10






  • 1




    $begingroup$
    Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
    $endgroup$
    – GEdgar
    Jan 3 at 15:34
















3












$begingroup$


Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.



It was a multiple choice questions with



a) $4sinh(3) sinh(1)$



b) $4sinh(1)sinh(3)$



I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .



Am I correct both option is correct ?
If wrong please explain why ?






complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    You're correct. Multiplication is commutative. The question was badly designed.
    $endgroup$
    – Michael Lugo
    Jan 3 at 15:10






  • 1




    $begingroup$
    Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
    $endgroup$
    – GEdgar
    Jan 3 at 15:34














3












3








3





$begingroup$


Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.



It was a multiple choice questions with



a) $4sinh(3) sinh(1)$



b) $4sinh(1)sinh(3)$



I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .



Am I correct both option is correct ?
If wrong please explain why ?






complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral







share|cite|improve this question











$endgroup$




Is $ sinh(x) sinh(y)=sinh(y) sinh(x)$? While evaluation a question on multiple integral I have got answer $4sinh(3) sinh(1)$.



It was a multiple choice questions with



a) $4sinh(3) sinh(1)$



b) $4sinh(1)sinh(3)$



I think both a and b option are correct since $sinh(1)$ ,$sinh(3)$ is multiplication of numbers it should commute but in answer option a is mention .



Am I correct both option is correct ?
If wrong please explain why ?






complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral




complex-numbers hyperbolic-geometry hyperbolic-functions multiple-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 3:23







sejy

















asked Jan 3 at 15:01









sejysejy

1579




1579








  • 2




    $begingroup$
    You're correct. Multiplication is commutative. The question was badly designed.
    $endgroup$
    – Michael Lugo
    Jan 3 at 15:10






  • 1




    $begingroup$
    Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
    $endgroup$
    – GEdgar
    Jan 3 at 15:34














  • 2




    $begingroup$
    You're correct. Multiplication is commutative. The question was badly designed.
    $endgroup$
    – Michael Lugo
    Jan 3 at 15:10






  • 1




    $begingroup$
    Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
    $endgroup$
    – GEdgar
    Jan 3 at 15:34








2




2




$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10




$begingroup$
You're correct. Multiplication is commutative. The question was badly designed.
$endgroup$
– Michael Lugo
Jan 3 at 15:10




1




1




$begingroup$
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
$endgroup$
– GEdgar
Jan 3 at 15:34




$begingroup$
Either that or it was a misprint or sejy misread it. From the information in the question, we cannot determine which of these three possibilities holds.
$endgroup$
– GEdgar
Jan 3 at 15:34










1 Answer
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$begingroup$

Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.






share|cite|improve this answer











$endgroup$













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    1 Answer
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    $begingroup$

    Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.






        share|cite|improve this answer











        $endgroup$



        Yes: $sinh:mathbb{C}to mathbb{C}$ so that $sinh(x),sinh(y)in mathbb{C}$. Thus, by commutativity of multiplication in $mathbb{C}$, $sinh(x)sinh(y)=sinh(y)sinh(x)$ for any $x,yin mathbb{C}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 15:19

























        answered Jan 3 at 15:10









        Antonios-Alexandros RobotisAntonios-Alexandros Robotis

        10.5k41641




        10.5k41641






























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