# Intuitive/geometric way to show that $\sinh x =-i\sin ix$ and $\cosh x=\cos ix$?

\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.

• Do you know Eulers identity? – imranfat Dec 2 at 23:57
• Yes I know Eulers identity – helpme Dec 3 at 0:00
• 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. – Ian Dec 3 at 0:10
• 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). – Gerry Myerson Dec 3 at 0:42
• I think @Ian means using $\sin z=(e^{iz}-e^{-iz})/(2i)$, or $e^{iz}=\cos z+i\sin z$. – Gerry Myerson Dec 3 at 11:58