Euclidean division exercise












6












$begingroup$


I faced the following problem in a previous abstract algebra session in my university:
Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






abstract-algebra polynomials







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    I faced the following problem in a previous abstract algebra session in my university:
    Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



    I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






    abstract-algebra polynomials







    share|cite|improve this question











    $endgroup$















      6












      6








      6





      $begingroup$


      I faced the following problem in a previous abstract algebra session in my university:
      Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



      I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






      abstract-algebra polynomials







      share|cite|improve this question











      $endgroup$




      I faced the following problem in a previous abstract algebra session in my university:
      Let $omega$ be a non-zero real number and n be a non-zero natural integer both supposed to be fixed. Calculate the remainder of the Euclidean division of the polynomial $(cos{omega}+Xsin{omega}) ^n$ by $X^2 +1$.



      I tried to expand $(cos{omega}+Xsin{omega}) ^n$ but it didn't look helpful for me. Can anyone help?






      abstract-algebra polynomials




      abstract-algebra polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 30 '18 at 21:53









      José Carlos Santos

      164k22131234




      164k22131234










      asked Dec 30 '18 at 21:23









      user7857462user7857462

      784




      784






















          2 Answers
          2






          active

          oldest

          votes


















          9












          $begingroup$

          Write the reminder $aX+b$. You have



          $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



          Substitute in this equation $X$ by $i$. You get
          $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



          $$e^{-in omega} = -ai+b$$



          Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            This is essentially the same answer as mathcounterexamples.net's answer, but it's a slightly different explanation and thought process, which seems to me to be slightly more natural (though I did upvote mce.net's answer). Hence I'll give my thought process on seeing this question.



            We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



            Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
            Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
              $endgroup$
              – Eric Towers
              Dec 31 '18 at 3:38










            • $begingroup$
              @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
              $endgroup$
              – jgon
              Dec 31 '18 at 3:57










            • $begingroup$
              Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
              $endgroup$
              – Eric Towers
              Dec 31 '18 at 4:05










            • $begingroup$
              @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
              $endgroup$
              – jgon
              Dec 31 '18 at 4:08












            • $begingroup$
              ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
              $endgroup$
              – Eric Towers
              Dec 31 '18 at 4:14













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            9












            $begingroup$

            Write the reminder $aX+b$. You have



            $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



            Substitute in this equation $X$ by $i$. You get
            $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



            $$e^{-in omega} = -ai+b$$



            Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






            share|cite|improve this answer









            $endgroup$


















              9












              $begingroup$

              Write the reminder $aX+b$. You have



              $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



              Substitute in this equation $X$ by $i$. You get
              $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



              $$e^{-in omega} = -ai+b$$



              Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






              share|cite|improve this answer









              $endgroup$
















                9












                9








                9





                $begingroup$

                Write the reminder $aX+b$. You have



                $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



                Substitute in this equation $X$ by $i$. You get
                $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



                $$e^{-in omega} = -ai+b$$



                Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.






                share|cite|improve this answer









                $endgroup$



                Write the reminder $aX+b$. You have



                $$P(X) = (cos omega + sinomega X)^n =Q(X)(X^2+1) +aX+b.$$



                Substitute in this equation $X$ by $i$. You get
                $$e^{in omega} = ai+b.$$ Substitute now $X$ by $-i$. You get



                $$e^{-in omega} = -ai+b$$



                Solving in $a,b$ you finally get $b =cos nomega$ and $a = sin nomega$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 30 '18 at 21:53









                mathcounterexamples.netmathcounterexamples.net

                26.9k22157




                26.9k22157























                    2












                    $begingroup$

                    This is essentially the same answer as mathcounterexamples.net's answer, but it's a slightly different explanation and thought process, which seems to me to be slightly more natural (though I did upvote mce.net's answer). Hence I'll give my thought process on seeing this question.



                    We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                    Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                    Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 3:38










                    • $begingroup$
                      @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 3:57










                    • $begingroup$
                      Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:05










                    • $begingroup$
                      @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 4:08












                    • $begingroup$
                      ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:14


















                    2












                    $begingroup$

                    This is essentially the same answer as mathcounterexamples.net's answer, but it's a slightly different explanation and thought process, which seems to me to be slightly more natural (though I did upvote mce.net's answer). Hence I'll give my thought process on seeing this question.



                    We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                    Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                    Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 3:38










                    • $begingroup$
                      @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 3:57










                    • $begingroup$
                      Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:05










                    • $begingroup$
                      @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 4:08












                    • $begingroup$
                      ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:14
















                    2












                    2








                    2





                    $begingroup$

                    This is essentially the same answer as mathcounterexamples.net's answer, but it's a slightly different explanation and thought process, which seems to me to be slightly more natural (though I did upvote mce.net's answer). Hence I'll give my thought process on seeing this question.



                    We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                    Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                    Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.






                    share|cite|improve this answer











                    $endgroup$



                    This is essentially the same answer as mathcounterexamples.net's answer, but it's a slightly different explanation and thought process, which seems to me to be slightly more natural (though I did upvote mce.net's answer). Hence I'll give my thought process on seeing this question.



                    We can calculate the remainder on division by a polynomial $fin k[x]$ by working in the ring $k[x]/(f)$, since the remainder of a polynomial $a$ on division by $f$ is the unique representative of $a$ in $k[x]/(f)$ of degree less than the degree of $f$.



                    Then recognize that $Bbb{R}[X]/(X^2+1)cong Bbb{C}$ via $Xmapsto i$.
                    Thus $(cos omega +Xsin omega)^nmapsto (e^{iomega})^n=e^{iomega n} = cos(nomega) + i sin(nomega)$. Thus the remainder on division by $X^2+1$ must be $cos(nomega)+Xsin(nomega)$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 31 '18 at 4:28

























                    answered Dec 30 '18 at 22:33









                    jgonjgon

                    14.8k22042




                    14.8k22042












                    • $begingroup$
                      Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 3:38










                    • $begingroup$
                      @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 3:57










                    • $begingroup$
                      Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:05










                    • $begingroup$
                      @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 4:08












                    • $begingroup$
                      ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:14




















                    • $begingroup$
                      Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 3:38










                    • $begingroup$
                      @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 3:57










                    • $begingroup$
                      Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:05










                    • $begingroup$
                      @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
                      $endgroup$
                      – jgon
                      Dec 31 '18 at 4:08












                    • $begingroup$
                      ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
                      $endgroup$
                      – Eric Towers
                      Dec 31 '18 at 4:14


















                    $begingroup$
                    Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
                    $endgroup$
                    – Eric Towers
                    Dec 31 '18 at 3:38




                    $begingroup$
                    Note that algebraic concerns alone cannot tell you whether you should use $X mapsto mathrm{i}$ or $X mapsto -mathrm{i}$ (since algebra doesn't distinguish Galois conjugates). mathcounterexamples.net does both.
                    $endgroup$
                    – Eric Towers
                    Dec 31 '18 at 3:38












                    $begingroup$
                    @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
                    $endgroup$
                    – jgon
                    Dec 31 '18 at 3:57




                    $begingroup$
                    @EricTowers That's what confuses me so much about mathcounterexamples.net's answer. There's absolutely no reason to do both, since it doesn't matter which isomorphism with $Bbb{C}$ you choose. You'll get the same answer either way.
                    $endgroup$
                    – jgon
                    Dec 31 '18 at 3:57












                    $begingroup$
                    Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
                    $endgroup$
                    – Eric Towers
                    Dec 31 '18 at 4:05




                    $begingroup$
                    Since algebra can't tell the difference, both specializations give (simultaneously) valid equations. We get an equation for $b+ mathrm{i} a$ and its conjugate $b - mathrm{i} a$, exactly as we should since the action taking $mathrm{i}$ to $-mathrm{i}$ is conjugation. (If the LHS had been real, the two equations would be redundant, again as expected, since $mathbb{R}$ is fixed by this action.)
                    $endgroup$
                    – Eric Towers
                    Dec 31 '18 at 4:05












                    $begingroup$
                    @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
                    $endgroup$
                    – jgon
                    Dec 31 '18 at 4:08






                    $begingroup$
                    @EricTowers, but you don't need both, since we've already fixed an isomorphism to $Bbb{C}$. The fact that $b+ia=e^{inomega}$ is already enough to figure out what $a$ and $b$ are, since we know how to expand $e^{inomega}$ in the $1,i$ basis for $Bbb{C}$.
                    $endgroup$
                    – jgon
                    Dec 31 '18 at 4:08














                    $begingroup$
                    ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
                    $endgroup$
                    – Eric Towers
                    Dec 31 '18 at 4:14






                    $begingroup$
                    ... or you've had $sin theta = frac{mathrm{e}^{mathrm{i}theta} - mathrm{e}^{-mathrm{i}theta}}{2mathrm{i}}$ beaten into you so hard that when you try both substitutions in your head you realize you can immediately solve for $a$ and $b$ in sines and cosines entirely algebraically, that is, without having to know an additional obscure analysis fact: Euler's identity. Notice: I'm not claiming mathcounterexample.net's answer is better or worse than yours. I'm only responding to your line "since frankly I don't follow several aspects ..." with how my head justified his manipulations.
                    $endgroup$
                    – Eric Towers
                    Dec 31 '18 at 4:14




















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