Question Number 113451 by pete last updated on 13/Sep/20
$$\mathrm{If}\:\mathrm{2x}=\mathrm{a}^{\mathrm{n}} +\mathrm{a}^{−\mathrm{n}} \:\mathrm{and}\:\mathrm{2y}=\mathrm{a}^{\mathrm{n}} −\mathrm{a}^{−\mathrm{n}} \:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}\:} \:\mathrm{in}\:\mathrm{its}\:\mathrm{simplest}\:\mathrm{form} \\ $$
Answered by bemath last updated on 13/Sep/20
$$\mathrm{4}{x}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}{n}} +{a}^{−\mathrm{2}{n}} +\mathrm{2} \\ $$$$\mathrm{4}{y}^{\mathrm{2}} =\:{a}^{\mathrm{2}{n}} +{a}^{−\mathrm{2}{n}} −\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:=\:\mathrm{1} \\ $$
Answered by Her_Majesty last updated on 13/Sep/20
$$\mathrm{2}{x}=\mathrm{2}{cosh}\left({nlna}\right) \\ $$$$\mathrm{2}{y}=\mathrm{2}{sinh}\left({nlna}\right) \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{1} \\ $$
Commented by Her_Majesty last updated on 13/Sep/20
$${you}'{re}\:{wrong}. \\ $$$$\mathrm{2}{x}={a}^{{n}} +{a}^{−{n}} ={e}^{{nlna}} +{e}^{−{nlna}} =\mathrm{2}×\left({e}^{{nlna}} +{e}^{−{nlna}} \right)/\mathrm{2}= \\ $$$$=\mathrm{2}{cosh}\left({nlna}\right) \\ $$$${similar}\:\mathrm{2}{y}=\mathrm{2}{sinh}\left({nlna}\right) \\ $$$${cosh}^{\mathrm{2}} \theta−{sinh}^{\mathrm{2}} \theta=\mathrm{1} \\ $$
Answered by MJS_new last updated on 13/Sep/20
$$\frac{\left({u}+{v}\right)^{\mathrm{2}} }{\mathrm{4}}−\frac{\left({u}−{v}\right)^{\mathrm{2}} }{\mathrm{4}}={uv} \\ $$$${here}\:{v}=\frac{\mathrm{1}}{{u}}\:\Rightarrow\:{uv}=\mathrm{1} \\ $$