# Thread: Hyperbolic functions and their values :(

1. ## 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

2. Hint:

$\displaystyle cosh^2(x) - sinh^2(x) =1$

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

I think you meant $\displaystyle \cosh^2(x) - \sinh^2(x)=1$.

4. Yes! you're right, thanks for the correction

5. When remembering the differences between the normal trigonometrical functions and their hyperbolic alternatives, I find it useful to use the,

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.

6. ## 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

7. $\displaystyle \cosh^2(x)-\sinh^2(x)=1$. 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 $\displaystyle \frac{\sinh(x)}{\cosh(x)}=\tanh(x)$.

8. 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

Editedpps I think you want the value of x anyway the others answer you

9. ## 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!