# Thread: Showing work for the natural number

1. ## Showing work for the natural number

Given that
$f(x) = \frac {e^x + e^{-x}}{2}$ and $g(x) = \frac {e^x - e^{-x}}{2}
$
show that $
[f(x)]^2 - [g(x)]^2 = 1$
Ok here is what I figured out so far. By pluging those 2 functions in to the equation and squaring it I get
$\frac {e^{2x} + e^{-2x}}{4} - \frac {e^{2x} + e^{-2x}}{4} = 1$.

Here's my question though. Wouldn't that make it equal $0$ once you subtract all the variables, as it would be $\frac {0}{4}$, which is equal to $0$.

2. $(\frac {e^{x} + e^{-x}}{2})^2\;=\; \frac {e^{2x} + e^{-2x} + 2}{4}$

3. Originally Posted by running-gag
$(\frac {e^{x} + e^{-x}}{2})^2\;=\; \frac {e^{2x} + e^{-2x} + 2}{4}$
Thanks. I finally got it now.