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Wolfram Alpha gives following results for $\cos(i x)$ and $\sin(i x)$ ,where $x\in\mathbb{R}$:

$\cos(ix)=\cosh(x)$

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

What is a reason why the first number is real while the second is complex ?

Definition of the $\cosh(x)$ and $\sinh(x)$ may be found here.

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Say that $f$ is real if $\overline{f(z)}=f(\overline z)$. Then, if $f$ is real and even, it's real on $i\mathbb R$, and if $f$ is real and odd, it's imaginary on $i\mathbb R$. - That's basically what Zev spelled out. –  Pierre-Yves Gaillard Oct 30 '11 at 15:58
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Any particular reason why you're using dotless i's for the imaginary unit? Just curious. –  Henning Makholm Oct 30 '11 at 16:00
    
@Henning,no special reason...you may edit and put $\mathrm{i}$ if you like so... –  pedja Oct 30 '11 at 16:06

2 Answers 2

up vote 5 down vote accepted

Note that, using Euler's formula, we can conclude that $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ and $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}.$$ Thus, $$\cos(ix)=\frac{e^{-x}+e^{x}}{2}=\cosh(x)$$ and $$\sin(ix)=\frac{e^{-x}-e^{x}}{2i}=i\left(\frac{e^{x}-e^{-x}}{2}\right)=i\sinh(x),$$ and it is clear from the definitions that $\sinh(x)$ and $\cosh(x)$ are both real for any $x\in\mathbb{R}$. Thus, $\cos(ix)$ is real for any $x\in\mathbb{R}$, and $\sin(ix)$ is imaginary for any $x\in\mathbb{R}$.

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It is directly related to the graphs of cosh and sinh: cosh is parabolic, while sinh drops off to negative y values when x is negative: This (-x,-y) portion of the sin graph could be taken as suggestive toward the role sin plays in representing imaginary numbers. Conversly the (-x,+y) values of cosh is representative of the role it plays as a carrier of positive/real numbers. If cosh had (-x,-y) values it would likely also carry imaginary components.

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