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## Revision as of 05:29, 25 August 2015

The hyperbolic sine function is defined by $$\mathrm{sinh}(z)=\dfrac{e^z-e^{-z}}{2}.$$

- Complex Sinh.jpg
Domain coloring of analytic continuation of $\sinh$.

## Contents

# Properties

## Theorem

The following formula holds:
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$
where $\sinh$ denotes the **hyperbolic sine** and $\cosh$ denotes the hyperbolic cosine.

## Proof

From the definition, $$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the hyperbolic cosine, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$ as was to be shown. █

## References

## Theorem

The following formula holds:
$$\sinh(az)=az {}_0F_1 \left( ; \dfrac{3}{2} ; \dfrac{(az)^2}{4} \right),$$
where $\sinh$ denotes the **hyperbolic sine** and ${}_0F_1$ denotes the hypergeometric pFq.

## Proof

## References

## Theorem

The Weierstrass factorization of **$\sinh(x)$** is
$$\sinh(x)=x\displaystyle\prod_{k=1}^{\infty} 1 + \dfrac{x^2}{k^2\pi^2}.$$

## Proof

## References

**Theorem:** The following formula holds:
$$\sinh(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^{2k+1}}{(2k+1)!}.$$

**Proof:** █

## Theorem

The following formula holds:
$$I_{\frac{1}{2}}(z)=\sqrt{\dfrac{2}{\pi z}}\sinh(z),$$
where $I_{\frac{1}{2}}$ denotes the modified Bessel function of the first kind and $\sinh$ denotes the **hyperbolic sine**.

## Proof

## References

## Theorem

The following formula holds:
$$\sin(z)=-i \sinh(iz),$$
where $\sin$ denotes the sine and $\sinh$ denotes the **hyperbolic sine**.

## Proof

From the definition of $\sin$ and $\sinh$ and the reciprocal of i, $$-i\sinh(iz) = \dfrac{e^{iz}-e^{-iz}}{2i} =\sin(z),$$ as was to be shown.

## References

## Theorem

The following formula holds:
$$\sinh(z)=-i\sin(iz),$$
where $\sinh$ is the **hyperbolic sine** and $\sin$ is the sine.

## Proof

By definition, $$\sinh(z) = \dfrac{e^{z}-e^{-z}}{2},$$ and so by the definition of $\sin$ and the reciprocal of i, we see $$-i\sinh(iz)=\dfrac{e^{iz}-e^{-iz}}{2i},$$ as was to be shown. █

## References

- 1964: Milton Abramowitz and Irene A. Stegun:
*Handbook of mathematical functions*... (previous) ... (next): $4.5.7$

**Hyperbolic trigonometric functions**