[SOLVED] Help simplify this hyperbolic trig function

**downthesun01** · February 22nd 2010, 10:46 PM

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.

**Prove It** · February 22nd 2010, 11:22 PM

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 $\cosh^2{t} - \sinh^2{t} = 1$ ...

**downthesun01** · February 22nd 2010, 11:26 PM

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..

**Prove It** · February 22nd 2010, 11:28 PM

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 $\sqrt{\sinh^2{t} \color{red}+\color{black}\cosh^2{t} + 1^2}$

**downthesun01** · February 22nd 2010, 11:36 PM

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,
$\sqrt{2cosh^2(t)}$
$<br /> \sqrt{2}cosh(t)$

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