What is the relation to $sinhx,coshx$ and $sinx,cosx$ [duplicate]

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  • What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?

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I've learned what $sinhx,coshx$ (the hyperbolic trig functions) are defined as formula, but how is it related to $sinx,cosx?$

The only thing I've noticed is that $cosh^2(x)-sinh^2(x)=1.$






trigonometry hyperbolic-functions







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  • See, for instance, "What's the intuition behind the identities $cos(z)=cosh(iz)$ and $sin(z)=-isinh(iz)$?
    – Blue
    Aug 31 at 1:30














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  • What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?

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I've learned what $sinhx,coshx$ (the hyperbolic trig functions) are defined as formula, but how is it related to $sinx,cosx?$

The only thing I've noticed is that $cosh^2(x)-sinh^2(x)=1.$






trigonometry hyperbolic-functions







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  • See, for instance, "What's the intuition behind the identities $cos(z)=cosh(iz)$ and $sin(z)=-isinh(iz)$?
    – Blue
    Aug 31 at 1:30












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This question already has an answer here:



  • What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?

    3 answers



I've learned what $sinhx,coshx$ (the hyperbolic trig functions) are defined as formula, but how is it related to $sinx,cosx?$

The only thing I've noticed is that $cosh^2(x)-sinh^2(x)=1.$






trigonometry hyperbolic-functions







share|cite|improve this question














This question already has an answer here:



  • What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?

    3 answers



I've learned what $sinhx,coshx$ (the hyperbolic trig functions) are defined as formula, but how is it related to $sinx,cosx?$

The only thing I've noticed is that $cosh^2(x)-sinh^2(x)=1.$





This question already has an answer here:



  • What's the intuition behind the identities $cos(z)= cosh(iz)$ and $sin(z)=-isinh(iz)$?

    3 answers






trigonometry hyperbolic-functions




trigonometry hyperbolic-functions






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asked Aug 31 at 1:27









Jason Kim

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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • See, for instance, "What's the intuition behind the identities $cos(z)=cosh(iz)$ and $sin(z)=-isinh(iz)$?
    – Blue
    Aug 31 at 1:30
















  • See, for instance, "What's the intuition behind the identities $cos(z)=cosh(iz)$ and $sin(z)=-isinh(iz)$?
    – Blue
    Aug 31 at 1:30















See, for instance, "What's the intuition behind the identities $cos(z)=cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Blue
Aug 31 at 1:30




See, for instance, "What's the intuition behind the identities $cos(z)=cosh(iz)$ and $sin(z)=-isinh(iz)$?
– Blue
Aug 31 at 1:30










1 Answer
1






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They’re related by Euler’s formula. Since $e^ix=cos x+i sin x$ we have $e^-ix=cos x-i sin x$. This reveals,



$$cosh (ix)=cos x$$



$$sinh (ix)=i sin x$$






share|cite|improve this answer



























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    They’re related by Euler’s formula. Since $e^ix=cos x+i sin x$ we have $e^-ix=cos x-i sin x$. This reveals,



    $$cosh (ix)=cos x$$



    $$sinh (ix)=i sin x$$






    share|cite|improve this answer
























      up vote
      1
      down vote



      accepted










      They’re related by Euler’s formula. Since $e^ix=cos x+i sin x$ we have $e^-ix=cos x-i sin x$. This reveals,



      $$cosh (ix)=cos x$$



      $$sinh (ix)=i sin x$$






      share|cite|improve this answer






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        They’re related by Euler’s formula. Since $e^ix=cos x+i sin x$ we have $e^-ix=cos x-i sin x$. This reveals,



        $$cosh (ix)=cos x$$



        $$sinh (ix)=i sin x$$






        share|cite|improve this answer












        They’re related by Euler’s formula. Since $e^ix=cos x+i sin x$ we have $e^-ix=cos x-i sin x$. This reveals,



        $$cosh (ix)=cos x$$



        $$sinh (ix)=i sin x$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 31 at 1:31









        Ahmed S. Attaalla

        14.1k11747




        14.1k11747












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