How does this trigonometric process work?

Question

I am currently learning about hyperbolic functions, and I have been presented with the following:$$\sinh^2 x = \frac{\cosh2 x - 1}{2}$$

This may seem like a simple conversion for some, and my apologies for my ignorance on the topic; however, I do not know where to start with this. If someone could please explain to me the process from "a" to "b" I would be extremely grateful.

Answer 1

The identity should be $$\sinh^2 x = \frac{\cosh 2x -1}{2}.$$

Probably the easiest way to show it is by changing to exponential form:

$$\sinh^2 x = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} -2 +e^{-2x}}{4} = \frac{(e^{2x}+e^{-2x})/2-1}{2} =\frac{\cosh 2x -1}{2}.$$

Answer 2

we have $$\sinh(x)=\frac{e^x-e^{-x}}{2}$$ and $$\cosh(x)=\frac{e^x+e^{-x}}{2}$$ plugging this in the term $$2\sinh(x)^2-\cosh(x)^2+1=\sinh(x)^2$$

Answer 3

In the question you made a small mistake. There is no $2$ in the denominator of its right hand side. By a small arithmetical confusion the $2$ of $\sinh$, $\cosh$ had gotten forcefully forward.