proving that the differences of squares of hyperbolic sin/cos is an integer.

Question

The hyperbolic sine and cosine are defined as following:

$\sinh(x)=\dfrac{e^x-e^{-x}}{2}$

$\cosh(x)=\dfrac{e^x+e^{-x}}{2}$

How do I show that their differences of squares are always an integer for all real numbers x?

hint appreciated!

Answer 1

Of course, there is the identity listed in the comments. But, for fun, let's see what we can do without that:

$$\cosh^2(x) = \frac{(e^{x}+e^{-x})^2}{4} = \frac{e^{2x}+e^{-2x}+2e^0}{4}$$

$$\sinh^2(x) = \frac{(e^{x}-e^{-x})^2}{4} = \frac{e^{2x}+e^{-2x}-2e^0}{4}$$

So: $$\begin{align}\require{cancel} \sinh^2(x) - \cosh^2(x) &= \frac{e^{2x}+e^{-2x}-2}{4} - \frac{e^{2x}+e^{-2x}+2}{4}\\ &=\frac{\cancel{e^{2x}}+e^{-2x}-2 - \cancel{e^{2x}}-e^{-2x}-2}{4}\\ &=\frac{\cancel{e^{-2x}}-2 -\cancel{e^{-2x}}-2}{4}\\ &=\frac{-2 -2}{4}\\ &=\frac{-4}{4}\\ &=-1\\ \end{align}$$

To prove for the other difference, we don't have to do near as much algebra: $$\begin{align} \cosh^2(x) - \sinh^2(x) &= -(\sinh^2(x) - \cosh^2(x)) \\ &= - (-1) \\ &= 1 \end{align}$$

As $1$ and $-1$ are integers, we have proven what we wanted.

Answer 2

Recall that that $$\begin{align}\cos iz&=\cosh z\\\sin iz&=i\sinh z\end{align}$$

Since $i^2=-1$, and $\cos^2z+\sin^2z=1$, we get that $$\cosh^2z-\sinh^2z=1$$