Express $e^x$ in terms of $\sinh x$

Question

Given $$ \sinh x = \frac{e^x-e^{-x}}{2} $$ how can I express $e^x$ in terms of $\sinh x$? I get $$ e^x = \sqrt{1 + 2e^x\sinh x} $$ where I can't get rid of $e^x$ on the right hand side.

Answer 1

If $e^x = y$, then $y-\frac 1y = 2 \sinh x$, followed by $y^2 - 2y \sinh x - 1 = 0$.

Using the quadratic formula, $$ e^x = y = \sinh x \pm \sqrt{(\sinh x)^2+1} $$ However, as $e^x>0$, the $-$ sign cannot happen. Hence, we have $$ e^x = \sinh x + \sqrt{(\sinh x)^2+1} $$

Answer 2

Since $\cosh x=(e^x+e^{-x})/2$ you can "cheat" and write $$e^x=\sinh x+\cosh x.$$ Also these obey rules similar to sines and cosines, except for signs, for example $\cosh^2 x-\sinh^2 x=1$, so $\cosh x=\sqrt{\sinh^2 x+1}$, so we also have $$ e^x=\sinh x + \sqrt{\sinh^2 x+1}. $$ Incidentally this gives us the formula $$ \sinh^{-1}y=\log\left(y+\sqrt{y^2+1}\right). $$