I've learned what $\sinh{x},\cosh{x}$ (the hyperbolic trig functions) are defined as formula, but how is it related to $\sin{x},\cos{x}?$
The only thing I've noticed is that $\cosh^2(x)-\sinh^2(x)=1.$
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Jason Kim
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See, for instance, "What's the intuition behind the identities $\cos(z)=\cosh(iz)$ and $\sin(z)=-i\sinh(iz)$? – Blue Aug 31 '18 at 01:30
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They’re related by Euler’s formula. Since $e^{ix}=\cos x+i \sin x$ we have $e^{-ix}=\cos x-i \sin x$. This reveals,
$$\cosh (ix)=\cos x$$
$$\sinh (ix)=i \sin x$$
Ahmed S. Attaalla
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