0

Struggling in Calc2, the question gives a value of sinh x = -3/4 and asking me to find the values of the remaining five hyperbolic functions. Can anybody help me as to how to approach this problem? I have the answers from the back of the book but I cannot understand how to answer it. Any help would be appreciated.

bankey
  • 331

1 Answers1

0

Use the identity $\cosh^2x-\sinh^2x \equiv 1$.

If $\sinh x = \frac{3}{4}$ then $$\cosh^2x - \left(\frac{3}{4}\right)^{\! 2} = 1$$ $$\cosh^2x - \left(\frac{9}{16}\right) = 1$$ $$\cosh^2x =\left(\frac{25}{16}\right)$$

It follows that $\cosh x = \pm\frac{1}{2}5/4$. Since $\cosh x \ge 1$ for all $x \in \mathbb{R}$ we have $\cosh x = \frac{1}{2}\sqrt{5/4}$.

Then, we find $\tanh(x)$ thanks to:

$\tanh(x)=\left(\frac{sinh(x)}{cosh(x)}\right)$.

And $sech(x)$ thanks to:

$sech(x)=\left(\frac{1}{cosh(x)}\right)$.

Again $cosech(x)$ with:

$cosech(x)=\left(\frac{1}{sinh(x)}\right)$.

  • The answer in the back of the book is costh x = 5/4. How did you get the +- 1/2 may I ask? And how would I be able to solve for functions such as tanh x, or sech x? – bankey Sep 05 '15 at 22:55
  • Thank you so much :) I may have some more questions on the homework problems that I did not get to but I have figured that one out that has been stumping me :) Thanks – bankey Sep 05 '15 at 23:05
  • No problem, a real pleasure.I'm trying to answer your comment by editing the answer. Learn these formulas BY HEART! – Revolucion for Monica Sep 05 '15 at 23:09
  • I don't know if you would know this, but if I may ask another question really quick: I need to rewrite the expression 2cosh(ln x) in terms of exponentials and simplify. If you can help that would be great, otherwise you already have done enough for me as it is :) The answer for that problem is x + 1/x but I don't get how to do that. I get that lnx is 1/x but I cant get the rest – bankey Sep 05 '15 at 23:12
  • @user268303 ha! Then you should ask another question, otherwise I can't code the answer in Latex, which is very handy and useful. – Revolucion for Monica Sep 05 '15 at 23:16
  • I will repost the question then. Sorry for being ignorant to using Latex, currently don't have the time to learn it haha. And is there any way to either instant message you or skype or something to ask questions? Unless you are busy of course, I don't expect you to help me any more if you're busy of course lol – bankey Sep 05 '15 at 23:20
  • @user268303 haha, It's 1h21 in my country I was only passing by ;) – Revolucion for Monica Sep 05 '15 at 23:21