$\mathrm{i}\sin(\varphi)=-\mathrm{i}\sin(\mathrm{i}z)$?

Question

The whole task is to proof this:

$$\sinh(z)=-\mathrm{i}\sin(\mathrm{i}z)$$

I used the definition of Euler-Formula und the Moivre-Theorem:

$$\begin{align*} \sinh(z)&=\frac{1}{2}(e^z-e^{-z}) \qquad \vert z=\varphi \mathrm{i}\\ &=\frac{1}{2}(e^{\mathrm{i}\varphi}-e^{-{\varphi\mathrm{i}}})\\ &=\frac{1}{2}(\cos(\varphi)+\mathrm{i}\sin(\varphi)-\cos(\varphi)+\mathrm{i}\sin(\varphi))\\ &=\mathrm{i}\sin(\varphi)\\ \end{align*}$$

And unfortunately the last (probably) easy steps are missing. Hints?

Answer 1

$$\sinh(z) = \dfrac{e^z-e^{-z}}{2} = -i\dfrac{e^{i^2z} - e^{-i^2z}}{2i} = -i\dfrac{e^{i(iz)} - e^{-i(iz)}}{2i} = -i\sin(iz).$$

Answer 2

Write down definitions:

$$\begin{cases}&i\sin z=i\cfrac{e^{iz}-e^{-iz}}{2i}=\cfrac{e^{iz}-e^{iz}}2=\sinh iz\\{}\\ &-i\sin iz=-i\cfrac{e^{-z}-e^z}{2i}=\cfrac{e^z-e^{-z}}2=\sinh z\end{cases}$$

So there is in one shot both your proof and also a disproof of the equality in your question's title.

Answer 3

Hint: it is easier to start from the right hand side of the equation.

$$-i\sin iz=-i \dfrac{1}{2i}(e^{-z}-e^z )$$ $$=\dfrac{e^z -e^{-z}}{2}=\sinh{z}$$

Answer 4

Use $z=(-i)iz$ and

$$\sinh z=\frac{e^z-e^{-z}}2=\frac{e^{(-i)iz}-e^{-(-i)iz}}2=-i\frac{e^{i\cdot iz}-e^{-i\cdot iz}}{2i}=-i\sin iz.$$