Does exponentiation distribute over the multiplication of trigoneomtric functions?












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I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?






algebra-precalculus trigonometry







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    $begingroup$
    Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 3:10
















0












$begingroup$


I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 3:10














0












0








0


1



$begingroup$


I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?






algebra-precalculus trigonometry







share|cite|improve this question











$endgroup$




I see exponentiation distributes over multiplication such as: $(3 . 4)^2 = (3^2 . 4^2)$, but I wonder: do the same rules hold for trigonometric functions? (i.e. is $(asin x)^2$ equal to $a^2sin^2x$)?






algebra-precalculus trigonometry




algebra-precalculus trigonometry






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edited Jan 8 at 4:29









naveen dankal

4,56921348




4,56921348










asked Jan 2 at 3:08









StartecStartec

1236




1236








  • 2




    $begingroup$
    Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 3:10














  • 2




    $begingroup$
    Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
    $endgroup$
    – Omnomnomnom
    Jan 2 at 3:10








2




2




$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10




$begingroup$
Are you sure about that? $(3 cdot 4)^2 = 12^2 = 144$, and $3^2 cdot 4^2 = 9 cdot 16 = 144$
$endgroup$
– Omnomnomnom
Jan 2 at 3:10










2 Answers
2






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oldest

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1












$begingroup$


Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.




$(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:



$$(a cdot b)^c = a^c cdot b^c$$



In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:



$$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$






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    1












    $begingroup$

    What sort of numbers are you dealing with? What sort of function space?



    In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:



    begin{equation}
    (a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
    end{equation}



    If it's endowed with associativity:



    begin{equation}
    (a*b)^n = a*b * a*b * a*b cdots * a*b
    end{equation}



    If it's also endowed with commutativity then:



    begin{align}
    (a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
    &= a^n * b^n = b^n * a^n
    end{align}



    I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.



    I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).



    If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.






    share|cite|improve this answer











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      2 Answers
      2






      active

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      2 Answers
      2






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      1












      $begingroup$


      Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.




      $(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:



      $$(a cdot b)^c = a^c cdot b^c$$



      In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:



      $$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$


        Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.




        $(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:



        $$(a cdot b)^c = a^c cdot b^c$$



        In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:



        $$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$


          Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.




          $(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:



          $$(a cdot b)^c = a^c cdot b^c$$



          In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:



          $$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$






          share|cite|improve this answer











          $endgroup$




          Edit: For the sake of posterity, the tenet about $(3 cdot 4)^2 = 3^2 cdot 4^2$ was from a previous version of OP's question. The core of my answer nonetheless answers the question, that being whether exponentiation distributes with trigonometric functions.




          $(3 cdot 4)^2$ does indeed equal $(3^2 cdot 4^2)$ (put both into a calculator: both are $144$), so what's the issue? Exponentiation does distribute over multiplication in the context of real numbers; that is to say:



          $$(a cdot b)^c = a^c cdot b^c$$



          In the case in the OP, note in particular, as a consequence of the commutativity of multiplication:



          $$(a cdot sin(x))^2 = (a cdot sin(x))cdot (a cdot sin(x)) = a cdot sin(x) cdot a cdot sin(x) = a cdot a cdot sin(x) cdot sin(x) = a^2 cdot sin^2(x)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 4:27







          user150203

















          answered Jan 2 at 3:12









          Eevee TrainerEevee Trainer

          8,88331640




          8,88331640























              1












              $begingroup$

              What sort of numbers are you dealing with? What sort of function space?



              In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:



              begin{equation}
              (a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
              end{equation}



              If it's endowed with associativity:



              begin{equation}
              (a*b)^n = a*b * a*b * a*b cdots * a*b
              end{equation}



              If it's also endowed with commutativity then:



              begin{align}
              (a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
              &= a^n * b^n = b^n * a^n
              end{align}



              I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.



              I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).



              If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                What sort of numbers are you dealing with? What sort of function space?



                In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:



                begin{equation}
                (a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
                end{equation}



                If it's endowed with associativity:



                begin{equation}
                (a*b)^n = a*b * a*b * a*b cdots * a*b
                end{equation}



                If it's also endowed with commutativity then:



                begin{align}
                (a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
                &= a^n * b^n = b^n * a^n
                end{align}



                I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.



                I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).



                If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  What sort of numbers are you dealing with? What sort of function space?



                  In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:



                  begin{equation}
                  (a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
                  end{equation}



                  If it's endowed with associativity:



                  begin{equation}
                  (a*b)^n = a*b * a*b * a*b cdots * a*b
                  end{equation}



                  If it's also endowed with commutativity then:



                  begin{align}
                  (a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
                  &= a^n * b^n = b^n * a^n
                  end{align}



                  I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.



                  I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).



                  If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.






                  share|cite|improve this answer











                  $endgroup$



                  What sort of numbers are you dealing with? What sort of function space?



                  In general, for $n in N$ over some set $(G,*)$ endowed with some closed operator $*$ then $forall a,b in G$:



                  begin{equation}
                  (a*b)^n = underbrace{left(a*bright) * (a*b) * (a*b) cdots * (a*b)}_{n mbox{ times}}
                  end{equation}



                  If it's endowed with associativity:



                  begin{equation}
                  (a*b)^n = a*b * a*b * a*b cdots * a*b
                  end{equation}



                  If it's also endowed with commutativity then:



                  begin{align}
                  (a*b)^n &= a*b * a*b * a*b cdots * a*b = underbrace{(a*a*cdots *a)}_{n mbox{ times}} * underbrace{(b*b*cdots *b)}_{ n mbox{ times}}\
                  &= a^n * b^n = b^n * a^n
                  end{align}



                  I would posit that you are working with the Field of Real Numbers which is a Commutative Field defined on multiplication and addition. Each operator $+$ and $*$ form Abelian Group. Abelian Groups are both associative and commutative.



                  I've added a lot of detail here not to try and overcomplicate but to let you know of different types of objects and operators. It's good to get an Abstract Algebra perspective on things early on as it will make your transition into Objecs such as Vectors, Matrices, https://en.wikipedia.org/wiki/Complex_number, etc much easier (as many do not satisfy this property in general).



                  If you're interested in any good intro to Abstract Algebra books, let me know. There is also some great free online video tutorials on the subject. I myself spent a huge amount of time on said videos this year and it's made my mathematics (in general) much more solid.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 7:34

























                  answered Jan 2 at 7:25







                  user150203





































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