# Thread: problem with finding values

1. ## problem with finding values

Hi
I have problem with the following:
Find the value of one of the six hyperbolic functions for x:
$\displaystyle cosh(x)=\frac{5}{4}$

P.S

2. Hello, Paymemoney!

I don't understand the question . . .

Find the value of one of the six hyperbolic functions for $\displaystyle x:$

. . $\displaystyle \cosh(x)=\frac{5}{4}\qquad{\color{blue}\Leftarrow\ ;\;\text{We have one!}}$

So what's the problem?
If they mean find another one . . .

$\displaystyle \text{Since }\,\cosh^2\!x - \sin^2\!x \:=\:1,\,\text{ then: }\:\sinh x \;=\;\pm\sqrt{\cosh^2\!h - 1}$

$\displaystyle \text{Hence: }\:\sinh x \;=\;\pm\sqrt{\left(\frac{5}{4}\right)^2-1} \;=\;\pm\sqrt{\frac{9}{16}}$

Therefore: .$\displaystyle \sinh x \;=\;\pm\frac{3}{4}$

3. Originally Posted by Soroban
Hello, Paymemoney!

I don't understand the question . . .

If they mean find another one . . .

$\displaystyle \text{Since }\,\cosh^2\!x - \sin^2\!x \:=\:1,\,\text{ then: }\:\sinh x \;=\;\pm\sqrt{\cosh^2\!h - 1}$

$\displaystyle \text{Hence: }\:\sinh x \;=\;\pm\sqrt{\left(\frac{5}{4}\right)^2-1} \;=\;\pm\sqrt{\frac{9}{16}}$

Therefore: .$\displaystyle \sinh x \;=\;\pm\frac{3}{4}$

yeh this is what i mean