`sinh^-1t = ln(t + sqrt(t^2 + 1))` Prove

Sunday, March 31, 2013

$\displaystyle{{\sinh}}^{{-{{1}}}}{t}={\ln{{\left({t}+\sqrt{{{{t}}^{{2}}+{1}}}\right)}}}$ Prove

To prove

$\displaystyle{{\sinh}}^{{-{{1}}}}{t}={\ln{{\left({t}+\sqrt{{{{t}}^{{2}}+{1}}}\right)}}}$

let

$\displaystyle{{\sinh}}^{{-{{1}}}}{t}={x}$

$\displaystyle{t}={\sinh{{\left({x}\right)}}}=\frac{{{{e}}^{{x}}-{{e}}^{{-{x}}}}}{{2}}=\frac{{{{e}}^{{x}}-{\left(\frac{{1}}{{{{e}}^{{x}}}}\right)}}}{{2}}=\frac{{{{e}}^{{{2}{x}}}-{1}}}{{{2}{\left({{e}}^{{x}}\right)}}}$

let $\displaystyle{t}=\frac{{{{e}}^{{{2}{x}}}-{1}}}{{{2}{\left({{e}}^{{x}}\right)}}}$

=> $\displaystyle{2}{{e}}^{{x}}{t}={{e}}^{{{2}{x}}}-{1}$

=> let $\displaystyle{{e}}^{{x}}={u}$ so,

$\displaystyle{2}{u}{t}={{u}}^{{2}}-{1}$

=> $\displaystyle{{u}}^{{2}}-{2}{u}{t}-{1}={0}$ is of the quadratic form $\displaystyle{a}{{x}}^{{2}}+{b}{x}+{c}={0}$ so finding the roots using the quadratic formula

$\displaystyle\frac{{-{b}\pm\sqrt{{{{b}}^{{2}}-{4}{a}{c}}}}}{{{2}{a}}}$

here in the equation $\displaystyle{{u}}^{{2}}-{2}{u}{t}-{1}={0}$

$\displaystyle{a}={1},{b}=-{2},{c}=-{1}$

$\displaystyle{u}=\frac{{-{\left(-{2}{t}\right)}\pm\sqrt{{{4}{{t}}^{{2}}-{4}{\left({1}\right)}{\left(-{1}\right)}}}}}{{2}}$

$\displaystyle{u}=\frac{{{2}{t}\pm\sqrt{{{4}{{t}}^{{2}}+{4}}}}}{{2}}$

= $\displaystyle\frac{{{2}{t}\pm{2}\sqrt{{{{t}}^{{2}}+{1}}}}}{{2}}$

= $\displaystyle{t}\pm\sqrt{{{{t}}^{{2}}+{1}}}$

Since $\displaystyle{u}={{e}}^{{x}}>{0}$ then $\displaystyle{t}+\sqrt{{{{t}}^{{2}}+{1}}}>{0}$

So $\displaystyle{{e}}^{{x}}={t}+\sqrt{{{{t}}^{{2}}+{1}}}$

$\displaystyle{x}={\ln{{\left({t}+\sqrt{{{{t}}^{{2}}+{1}}}\right)}}}$

Since

$\displaystyle{{\sinh}}^{{-{{1}}}}{t}={x}$

it follows that

$\displaystyle{{\sinh}}^{{-{{1}}}}{t}={\ln{{\left({t}+\sqrt{{{{t}}^{{2}}+{1}}}\right)}}}$

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)