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I would like to understand the relationship betwene $e^{i\cdot \theta}$ and hyperbolic sine and cosine. Here is what I have done so far:

Given: $$\sinh(x)+\cosh(x)=e^x $$ $$i\sin(\theta)+\cos(\theta)=e^{i\theta} $$

Replaced $x$ in the first equation with $i\theta$ :

$$\sinh(i\theta)+\cosh({i\theta})=e^{i\theta} $$

Given: $$i\sinh(x)= \sin(ix)$$ $$\cosh(x)=\cos(ix) $$ Replaced $x$ with $i\theta$ $$i\sinh({i\theta})= \sin(i\cdot{i\theta})=\sin(-\theta)$$ $$\cosh({i\theta})=\cos(i\cdot {i\theta})=\cos(-\theta) $$

$$\sinh(x)= \frac{\sin(ix)}{i}$$

I feel like I'm doing meaningless symbolic manipulation that likely is flawed on some level.

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  • $\begingroup$ It only feels meaningless until you realise that all these functions are defined on $\mathbb{C}$, and not only on $\mathbb{R}$. Then you understand that these are fundamental identities. $\endgroup$ – Daniel Fischer Oct 16 '13 at 18:30
  • $\begingroup$ You do know that $\sinh x = \frac12(e^x - e^{-x})$, right? $\endgroup$ – MJD Oct 16 '13 at 18:34
  • $\begingroup$ Now replace $x$ with $-i \theta$ and add the resulting equations. Don't forget that sine is odd and cosine is even! $\endgroup$ – The Chaz 2.0 Oct 16 '13 at 18:34
  • $\begingroup$ @DanielFischer Thankyou for the encouragement. $\endgroup$ – User3910 Oct 16 '13 at 18:34
  • $\begingroup$ @TheChaz2.0 I don't quite follow. $\endgroup$ – User3910 Oct 16 '13 at 18:47
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You could have obtained these identites a little easier using the definitions $$\begin{align*} \sin(x) & = \frac 1 {2i} (e^{ix} - e^{-ix}) \\ \cos(x) & = \frac 1 2 (e^{ix} + e^{-ix}) \\ \sinh(x) & = \frac 1 2 (e^x - e^{-x}) \\ \cosh(x) & = \frac 1 2 (e^x + e^{-x}) \end{align*}$$

As @DanielFischer said, these make much more sense when considered as whole functions on $\mathbb C$.

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  • $\begingroup$ Does $x=\theta$ here? $\endgroup$ – User3910 Oct 16 '13 at 18:37
  • $\begingroup$ @JoeHobbit $x\in\mathbb C$ is just any variable. It can even be $\text{car}$, but noone wants to write $$\sin(\text{car})$$ ;-) $\endgroup$ – AlexR Oct 16 '13 at 18:38
  • $\begingroup$ @AlexR That's indeed not a good name for a variable. Now $\sin (\text{bad})$, that makes sense. $\endgroup$ – Daniel Fischer Oct 16 '13 at 18:42
  • $\begingroup$ @DanielFischer and $\sin(\text{good}) \approx \sin(\text{food})$? $\endgroup$ – AlexR Oct 16 '13 at 18:43
  • $\begingroup$ It seems you missed the pun. You're not a seafarer, are you? $\endgroup$ – Daniel Fischer Oct 16 '13 at 18:45

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