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 $\cosh x = \frac{1}{2}\sqrt{13}$.
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Bumbble Comm
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The trick is....
$$\cosh^2x-\sinh^2x=1$$
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Bumbble Comm
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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)) $
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user227676
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Do you want to say that $cosh^2x-sinh^2x=1$? Yes that is correct because of this:
$1/4[e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}]= 1/4 \times 4=1$
0
Bumbble Comm
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Most books I have seen use this way to solve this problem using the same identity mentioned in the accepted answer:
$$\sqrt{1+\sinh^2x}=\cosh x$$
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 $\cosh x = \frac{1}{2}\sqrt{13}$.