Given the value of sinh(x)

for example sinh(x) = 3/2

How can I calculate the value of cosh(x) ?

  • 11
    $\begingroup$ Use identity: $\cosh^2(x)-\sinh^2(x)=1$ $\endgroup$
    – alans
    Jul 11, 2014 at 18:47
  • 5
    $\begingroup$ @alans When you hover your mouse over "add comment" you will see the following: Use comments to ask for more information or to suggest improvements. Avoid answering questions in comments. $\endgroup$ Jul 11, 2014 at 19:26

5 Answers 5


Use the identity $\cosh^2x-\sinh^2x \equiv 1$. If $\sinh x = \frac{3}{2}$ then $$\cosh^2x - \left(\frac{3}{2}\right)^{\! 2} = 1$$ $$\cosh^2x - \frac{9}{4} = 1$$ $$\cosh^2x = \frac{13}{4}$$ It follows that $\cosh x = \pm\frac{1}{2}\sqrt{13}$. Since $\cosh x \ge 1$ for all $x \in \mathbb{R}$ we have $\cosh x = \frac{1}{2}\sqrt{13}$.

  • $\begingroup$ Please explain the downvote. $\endgroup$ Jul 11, 2014 at 18:51
  • $\begingroup$ You made a sign error. $\sinh$ maps $\mathbb{R}$ onto itself. $\endgroup$
    – Ian
    Jul 11, 2014 at 18:51
  • $\begingroup$ I think you want $\sinh x = \frac{3}{2}$ in your premise, not $\sinh^2$. (I tried to edit but it wasn't substantial enough since it was only two characters.) $\endgroup$ Jul 11, 2014 at 20:11
  • $\begingroup$ @IsmailBadawi You should not edit other people's posts. $\endgroup$ Jul 11, 2014 at 20:52
  • 8
    $\begingroup$ @FlybyNight Stack Exchange disagrees with you: meta.stackexchange.com/questions/120576/… $\endgroup$ Jul 11, 2014 at 21:15

The trick is....



Most books I have seen use this way to solve this problem using the same identity mentioned in the accepted answer: $$\sqrt{1+\sinh^2x}=\cosh x$$

$\therefore \sqrt{1+(\frac{3}{2})^2}=\cosh x$

$ \frac{\sqrt{13}}{2}=\cosh x$


Using $\cosh^2x-\sinh^2x=1$ you can evaluate it.

But unlike circular trig functions, there is only a single value for

$ \cosh( \sinh^{-1}(x)) $


Do you want to say that $cosh^2x-sinh^2x=1$? Yes that is correct because of this: $1/4[e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}]= 1/4 \times 4=1$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .