Question

If tanhx = 12/13, find the values of the other hyperbolic functions at x.

Answer 1

\[\tanh x = \frac{\sinh x}{\cosh x}\]

Answer 2

\[\frac{ 1 }{ coshx } = \sqrt{1-\tanh^2x}\]

Answer 3

\[sinhx = \sqrt{\cosh^2x-1}\]

Answer 4

Did I misunderstand the question?

Answer 5

Then you plug in the values and now you have to find the reciprocals.

Answer 6

Wait.

Answer 7

\[\tanh x = \frac{\sinh x}{\cosh x} = \frac{12}{13}\]

Answer 8

yeah

Answer 9

But that doesn't necessarily mean that \(\sinh(x)=12\)
That only give you the proportion.

Answer 10

So use \[
(13c)^2-(12c)^2=1
\]and solve for \(c\) I guess.

Answer 11

I knew it was too good to be true. Carry on. :D

Answer 12

It's too tedious

Answer 13

Well consider if \(\tan(x)=2/5\). Clearly \(\sin(x)\neq 2\) since it never gets that high.

Answer 14

Yeah coshx = 13/5

Answer 15

http://www.wolframalpha.com/input/?i=%2813c%29%5E2%E2%88%92%2812c%29%5E2%3D1

This tells us \(c=\pm 1/5\) So:\[
\sinh(x)=\pm \frac {12}{5}\quad \cosh(x)=\pm \frac {13}{5}
\]Both are positive or both are negative.

Answer 16

Would have to know the quadrant to tell.

Answer 17

So I should add the + or - sign ?

Answer 18

The reciprocals should be easy then, yeah?

Answer 19

Make it only positive, then say another possibility is that both are negative.

Answer 20

alright

Answer 21

Or you could be really cute and say \[
|\cosh (x)|=\frac {13}{5}
\]

Answer 22

hahahaha!

Answer 23

Technically you can't get any more accurate.

Answer 24

Wait how did you figure out what coshx and sinhx were o_O

Answer 25

Started with:\[\tanh x = \frac{\sinh x}{\cosh x} = \frac{12}{13}\]
So I knew\[\sinh x=12c\quad \cosh x=13c\]

Answer 26

I used the property: \[
\cosh^2x-\sinh^2 x=1
\]To get\[
(13c)^2−(12c)^2=1
\]

Answer 27

Found out that \(|c|=1/5\)

Answer 28

Ah, wow your way is much better than mine, and much simpler.