starting from the definition of sinhx, show that when x is real the real value of sinh^-1x is given by sin^-1(x)=ln((x+root(x^2+1))
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starting from the definition of sinhx, show that when x is real the real value of sinh^-1x is given by sin^-1(x)=ln((x+root(x^2+1))
$\displaystyle \displaystyle \begin{align*} y &= \sinh^{-1}{x} \\ \sinh{y} &= x \\ \frac{e^y - e^{-y}}{2} &= x \\ e^y - e^{-y} &= 2x \\ e^{2y} - 1 &= 2x\,e^y \\ e^{2y} - 2x\,e^y &= 1 \\ e^{2y} - 2x\,e^y + \left(-x\right)^2 &= \left(-x\right)^2 + 1 \\ \left( e^y - x \right)^2 &= x^2 + 1 \\ e^y - x &= \pm \sqrt{x^2 + 1} \\ e^y &= x \pm \sqrt{x^2 + 1} \\ y &= \ln{\left( x \pm \sqrt{x^2 + 1} \right)} \end{align*}$Quote:
Originally Posted by Magical
But of course, this is only defined for positive values of $\displaystyle \displaystyle \begin{align*} x \pm \sqrt{x^2 + 1} \end{align*}$, and since $\displaystyle \displaystyle \begin{align*} \sqrt{x^2 + 1} > x \end{align*}$ for all real $\displaystyle \displaystyle \begin{align*} x \end{align*}$, that means
$\displaystyle \displaystyle \begin{align*} \sinh^{-1}{x} \equiv \ln{\left( x + \sqrt{x^2 + 1} \right)} \end{align*}$
