# Math Help - [SOLVED] Help simplify this hyperbolic trig function

1. ## [SOLVED] Help simplify this hyperbolic trig function

This should be really easy, but I'm not getting it for whatever reason

$\sqrt{sinh^2(t) - cosh^2(t) + 1^2}$

The answer should be $\sqrt{2}cosh(t)$ But I don't understand how. There must be some simple identity that I'm missing.

2. Originally Posted by downthesun01
This should be really easy, but I'm not getting it for whatever reason

$\sqrt{sinh^2(t) - cosh^2(t) + 1^2}$

The answer should be $\sqrt{2}cosh(t)$ But I don't understand how. There must be some simple identity that I'm missing.
Remember that $\cosh^2{t} - \sinh^2{t} = 1$...

3. Originally Posted by Prove It
Remember that $\cosh^2{t} - \sinh^2{t} = 1$...
Yeah, I realize that but you're going to have to dumb down your point a bit for me haha..

4. Originally Posted by downthesun01
Yeah, I realize that but you're giving to have to dumb down your point a bit for me haha..
This means $\sinh^2{t} = \cosh^2{t} - 1$.

Substitute it in and simplify...

Upon further inspection, in order to obtain the answer of $\sqrt{2}\cosh{t}$, the original expression needs to be $\sqrt{\sinh^2{t} \color{red}+\color{black}\cosh^2{t} + 1^2}$

5. I figured out the problem. The original expression is
$\sqrt{sinh^2(t) + (-cosh(t))^2 + 1^2}$

Which is the same as $\sqrt{sinh^2(t) + cosh^2(t) + 1^2}$
So,
$\sqrt{2cosh^2(t)}$
$
\sqrt{2}cosh(t)$