problem with finding values

**Paymemoney** · January 8th 2010, 03:17 PM

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

P.S

**Soroban** · January 8th 2010, 04:02 PM

Hello, Paymemoney!

I don't understand the question . . .

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

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

So what's the problem?

If they mean find another one . . .

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

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

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

**Paymemoney** · January 8th 2010, 04:31 PM

Originally Posted by Soroban

Hello, Paymemoney!

I don't understand the question . . .

If they mean find another one . . .

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

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

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

yeh this is what i mean

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