# Hyperbolic functions and their values :(

• Jun 9th 2009, 11:41 PM
klooless
Hyperbolic functions and their values :(
Some help in the right direction? I have no idea where to start:

Question:
If sinh x = 3/4, find the values of the other hyperbolic functions at x.

I would greatly appreciate any help! I'm taking a distance course and learning all this by yourself is tough!!

cheers,

klooless
• Jun 10th 2009, 12:39 AM
Ruun
Hint:

$\displaystyle cosh^2(x) - sinh^2(x) =1$
• Jun 10th 2009, 12:43 AM
flyingsquirrel
Quote:

Originally Posted by Ruun
$\displaystyle sinh^2(x) - cosh^2(x)=1$

(Surprised)

I think you meant $\displaystyle \cosh^2(x) - \sinh^2(x)=1$.
• Jun 10th 2009, 12:51 AM
Ruun
Yes! you're right, thanks for the correction
• Jun 10th 2009, 12:56 AM
craig
When remembering the differences between the normal trigonometrical functions and their hyperbolic alternatives, I find it useful to use the,

Quote:

2 sinhs change the sign.
That is, if there are two sinhs multiplied together, as in the $\displaystyle sin^2$, the you change the sign that you would use normally, in this case it is a minus instead of a plus.
• Jun 11th 2009, 09:10 AM
klooless
Still having trouble; hyperbolic functions...
Thank you to Ruun, flyingsquirrel and Craig for giving me tips to solve;

Question:
If sinh x = 3/4, find the values of the other hyperbolic functions at x.

I have tried solving it using terms of the e^x definitions, and also working through cosh^2 (x) - sinh^2 (x) = 1, but I am not getting anywhere. I'm not sure what to do now, or how much help anyone can give me...

Thanks for reading and posting earlier, any further advice would be wonderful!

cheers
• Jun 11th 2009, 09:55 AM
Ruun
$\displaystyle \cosh^2(x)-\sinh^2(x)=1$. It is given that $\displaystyle \sinh(x)=\frac{3}{4}$, plugging in, into the first equation, will give you the $\displaystyle \cosh(x)$. The last thing you have to known, is that, as an equivalent of Euclidean trigonometry $\displaystyle \frac{\sinh(x)}{\cosh(x)}=\tanh(x)$.
• Jun 11th 2009, 10:03 AM
Amer
Quote:

Originally Posted by klooless
Thank you to Ruun, flyingsquirrel and Craig for giving me tips to solve;

Question:
If sinh x = 3/4, find the values of the other hyperbolic functions at x.

I have tried solving it using terms of the e^x definitions, and also working through cosh^2 (x) - sinh^2 (x) = 1, but I am not getting anywhere. I'm not sure what to do now, or how much help anyone can give me...

Thanks for reading and posting earlier, any further advice would be wonderful!

cheers

$\displaystyle sinhx= \frac{3}{4}$

$\displaystyle \frac{e^x - e^{-x} }{2} = \frac{3}{4}$

$\displaystyle e^x - e^{-x} = \frac{3}{2}$

$\displaystyle e^{2x} - 1 = \frac{3e^x}{2}$ by multiply with $\displaystyle e^x$

$\displaystyle e^{2x} -\frac{3e^x}{2} - 1 = 0$

let t= $\displaystyle e^x$

$\displaystyle t^2 - \frac{3t}{2} - 1 = 0$

$\displaystyle t=\frac{ \frac{3}{2} \pm \sqrt{ \frac{9}{4} - 4(1)(-1) } } {2}$

$\displaystyle t=\frac{ \frac{3}{2} \pm \sqrt{ \frac{25}{4} } } {2}$

$\displaystyle e^x = \frac{\frac{3}{2} + \frac{5}{2} } {2}$

or

$\displaystyle e^x = \frac{\frac{3}{2} - \frac{5}{2} } {2}$

I think it clear right

Edited:opps I think you want the value of x anyway the others answer you
• Jun 11th 2009, 10:41 AM
klooless
Thank you :ddd
Runn - that really cleared it up nicely for me, thanks for spelling it out!

Amer - Super clear explanation, not exactly what this question asked for, but it helped with some other problems! Thanks!

Cheers! :)