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Can someone show me:

If $x$ is a real number, then $\cos^2(x)+\sin^2(x)= 1$.

Is it true that $\cos^2(z)+\sin^2(z)=1$, where $z$ is a complex variable?

Note :look [this ] in wolfram alpha showed that's true !!!!

Thank you for your help

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    $\begingroup$ How do you define $\cos$ and $\sin$ on $\mathbb{C}$ ? $\endgroup$ Jun 14, 2015 at 23:30
  • $\begingroup$ In R ( real number) but z is a complex variable $\endgroup$ Jun 14, 2015 at 23:31
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    $\begingroup$ @salimmath15: That doesn't sound like a definition. Say, if claim that $\sin i=42$, how would you argue that I'm wrong? $\endgroup$ Jun 14, 2015 at 23:35
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    $\begingroup$ The LHS and RHS are both entire functions on $\mathbb{C}$. Since they coincides over $\mathbb{R}$, the coincides over the whole $\mathbb{C}$. $\endgroup$ Jun 14, 2015 at 23:35
  • $\begingroup$ look :wolframalpha.com/input/… $\endgroup$ Jun 14, 2015 at 23:35

3 Answers 3

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$$ \cos^2z+\sin^2z=(\cos z+i\sin z)(\cos z-i\sin z)=e^{iz}e^{-iz}=e^{iz-iz}=e^0=1. $$

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You can use the identity theorem. As they are just sums of exponentials, $\sin(z)$ and $\cos(z)$ are holomorphic, and on the real axis $\sin^2(x)+\cos^2(x)=1$. As $\mathbb{R}$ is a set with an accumulation point (namely any point in $\mathbb{R}$), they agree everywhere.

Mercy's answer is a bit simpler, but this is a good principle to keep in mind when trying to show other identities that are true for real numbers.

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  • $\begingroup$ :The answer of mercy is enough and clear at all and for all $\endgroup$ Jun 14, 2015 at 23:44
  • $\begingroup$ @ Andrew ledesma , sorry your answer is very complicated and it's not helpful $\endgroup$ Jun 14, 2015 at 23:54
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    $\begingroup$ @salimmath15 his answer is not complicated and it is very helpful for you as you learn more about complex variables. That is part of the beauty of this website. You don't just get answers at your level, you get a spectrum of understanding. $\endgroup$ Jun 15, 2015 at 0:17
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    $\begingroup$ @salinmath15 actually if you want to learn complex analysis this is something you want to keep in mind. This answer is not complicated at all and it is very helpful, perhaps not for you. $\endgroup$ Jun 15, 2015 at 0:42
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    $\begingroup$ @salimmath15 I would strongly suggest that you have a look at the link that was provided herein. This is a robust answer and ought to be quite useful to many others. $\endgroup$
    – Mark Viola
    Jun 15, 2015 at 3:55
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by definition (In complex Analysis):

cosz= ((e^iz)+(e^-iz))/2

sinz= ((e^iz)-(e^-iz))/2

L.H.S=(cosz+isinz)(cosz-sinz)

substitute with definitions above and manipulate algebraically. You should get this at the end: (e^iz)(e^-iz)= e^0=1= R.H.S

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