# Calculate cosh(x) given sinh(x)

Given the value of sinh(x)

for example sinh(x) = 3/2

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

• Use identity: $\cosh^2(x)-\sinh^2(x)=1$ Jul 11, 2014 at 18:47

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 $\cosh x = \frac{1}{2}\sqrt{13}$.

• Please explain the downvote. Jul 11, 2014 at 18:51
• You made a sign error. $\sinh$ maps $\mathbb{R}$ onto itself.
– Ian
Jul 11, 2014 at 18:51
• 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.) Jul 11, 2014 at 20:11
• @IsmailBadawi You should not edit other people's posts. Jul 11, 2014 at 20:52
• @FlybyNight Stack Exchange disagrees with you: meta.stackexchange.com/questions/120576/… Jul 11, 2014 at 21:15

The trick is....

$$\cosh^2x-\sinh^2x=1$$

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: $1/4[e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}]= 1/4 \times 4=1$