Question Number 190935 by Spillover last updated on 14/Apr/23
$$\mathrm{G}{iven}\:{that}\:\:\:\:\:\:\:\:\mathrm{sinh}\:^{−\mathrm{1}} {y}=\mathrm{sech}\:^{−\mathrm{1}} {y}\:\:\:\: \\ $$$${show}\:{that}\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:\:\: \\ $$$$ \\ $$
Answered by mehdee42 last updated on 14/Apr/23
$${sinh}^{−\mathrm{1}} {y}={a}\Rightarrow{y}={sinha}=\frac{{e}^{{a}} −{e}^{−{a}} }{\mathrm{2}} \\ $$$${sech}^{−\mathrm{1}} {y}={a}\Rightarrow{y}={secha}=\frac{\mathrm{2}}{{e}^{{a}} +{e}^{−{a}} } \\ $$$$\frac{{e}^{{a}} −{e}^{−{a}} }{\mathrm{2}}=\frac{\mathrm{2}}{{e}^{{a}} +{e}^{−{a}} }\Rightarrow{e}^{\mathrm{2}{a}} −{e}^{−\mathrm{2}{a}} =\mathrm{4} \\ $$$${if}\:\:{e}^{\mathrm{2}{a}} ={u}\Rightarrow{u}^{\mathrm{2}} −{u}−\mathrm{1}=\mathrm{0}\Rightarrow{u}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}={e}^{\mathrm{2}{a}} \: \\ $$$${y}^{\mathrm{2}} =\frac{{e}^{\mathrm{2}{a}} +{e}^{−\mathrm{2}{a}} −\mathrm{2}}{\mathrm{4}}=\frac{\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{2}}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by Spillover last updated on 15/Apr/23
$$\mathrm{thank}\:\mathrm{you} \\ $$