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Using hyperbolic trigonometric function identities is there a way to prove the following equation?

$$\frac{1}{2} (\cosh(2x)-1) = \sinh^2(x)$$

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3 Answers

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We have

$$\frac12\left(\cosh(2x)-1\right)=\frac14(e^{2x}+e^{-2x}-2)=\frac14(e^x-e^{-x})^2=\sinh^2 x$$

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This follows immediately from the corresponding identity

$$\frac12(\cos 2x - 1) = -\sin^2 x$$

for circular functions if you use the facts that $\cos ix = \cosh x$ and $\sin ix = i\sinh x$, provided that you accept or prove this identity instead.

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It holds:

$cosh(x\pm y)=cosh(x)cosh(y)\pm sinh(x)sinh(y)$

With x=y we obtain:

$cosh(2x)=cosh(x)^2+ sinh(x)^2$

With the usage of $cosh^2(x)-sinh^2(x)=1$ we obtain:

$cosh(2x)=1+sinh^2(x) + sinh(x)^2=1+2sinh^2(x) \iff \frac{1}{2}(cosh^2(x)-1)=sinh^2(x)$

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