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$$\begin{align} \sinh(x)&=-i\sin(ix) \\ \cosh(x)&=\phantom{-i}\cos(ix) \end{align}$$

Why are these identities true?

Other than using algebra and formulas, is there a more intuitive/geometric way to show that the above is true?

Attempt
The equation of a unit circle is: $x^2+y^2=1$
The equation of a unit hyperbola is: $x^2-y^2=1$
If $y\rightarrow iy$, then a circle will become a hyperbola.

For a circle: $y=\sin(θ)$
For a hyperbola: $iy=\sinh(θ)$ ---> $y=-i\sinh(θ)$

For a circle: $x=\cos(θ)$
For a hyperbola: $x=\cosh(θ)$

But obviously $\cos(x)=\cosh(x)$ and $\sin(x)=-i\sinh(x)$ are wrong.

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  • $\begingroup$ Do you know Eulers identity? $\endgroup$ – imranfat Dec 2 at 23:57
  • $\begingroup$ Yes I know Eulers identity $\endgroup$ – helpme Dec 3 at 0:00
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    $\begingroup$ I'm not sure you're going to get much geometric intuition for what $\sin(ix)$ should even mean in the first place, without passing through its connection to the exponential. $\endgroup$ – Ian Dec 3 at 0:10
  • $\begingroup$ The geometric definition of the hyperbolic functions is a little trickier than just going $y=i\sinh\theta$. See the diagram at en.wikipedia.org/wiki/Hyperbolic_function (but I agree with @Ian about the likelihood of success of a purely geometric argument). $\endgroup$ – Gerry Myerson Dec 3 at 0:42
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    $\begingroup$ I think @Ian means using $\sin z=(e^{iz}-e^{-iz})/(2i)$, or $e^{iz}=\cos z+i\sin z$. $\endgroup$ – Gerry Myerson Dec 3 at 11:58

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