1.) Show that $\displaystyle \cosh ^2 x - \sinh ^2 x = 1$ for all $\displaystyle x$
2.) Find all solutions of $\displaystyle \sinh (x^2 - 1) = 0$
Thanks again guys.
use the fact that $\displaystyle \sinh x = \frac {e^x - e^{-x}}2$ and $\displaystyle \cosh x = \frac {e^x + e^{-x}}2$
just set $\displaystyle x^2 - 1 = 0$2.) Find all solutions of $\displaystyle \sinh (x^2 - 1) = 0$
to see why this is a solution (that is, why the solution to $\displaystyle \sinh x = 0$ is $\displaystyle x = 0$) you can use the definition above
$\displaystyle \cosh^2 x - \sinh^2 x = \left( \frac {e^x + e^{-x}}2 \right)^2 - \left( \frac {e^x - e^{-x}}2 \right)^2 = ...$
if you are familiar with complex numbers, there is another approach...but this is pretty straight forward, so this is good enough. plus, it gives you more stuff to write so you can show of your elegance in writing math
that is only one solution, there are two
And for #2, I came out with $\displaystyle x=1$ ....does that look correct?