Question Number 79735 by Rio Michael last updated on 27/Jan/20
$${write}\:{tanhx}\:{in}\:{terms}\:{of}\:{e},\:{hence}\:{prove}\:{that}\: \\ $$$${tanh}\mathrm{2}{x}\:=\:\frac{\mathrm{2}{tanhx}}{\mathrm{1}+{tanh}^{\mathrm{2}} {x}} \\ $$
Answered by Henri Boucatchou last updated on 28/Jan/20
$${tanhx}=\frac{{e}^{{x}} −{e}^{−{x}} }{{e}^{{x}} +{e}^{−{x}} }\: \\ $$$$\:\:{tanh}\mathrm{2}{x}=\frac{{sh}\mathrm{2}{x}}{{ch}\mathrm{2}{x}}=\frac{\mathrm{2}{shx}\:{chx}}{{ch}^{\mathrm{2}} {x}+{sh}^{\mathrm{2}} {x}} \\ $$$${we}\:\:{have}\:\:\begin{cases}{{sh}\mathrm{2}{x}=\mathrm{2}{shx}\:{chx}}\\{{ch}\mathrm{2}{x}={ch}^{\mathrm{2}} {x}+{sh}^{\mathrm{2}} {x}}\end{cases}\:\Rightarrow\:\:\boldsymbol{{th}}\mathrm{2}\boldsymbol{{x}}=\frac{\boldsymbol{{sh}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{ch}}\mathrm{2}\boldsymbol{{x}}}=\frac{\mathrm{2}\boldsymbol{{shx}}\:\boldsymbol{{chx}}}{\boldsymbol{{ch}}^{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{{sh}}^{\mathrm{2}} \boldsymbol{{x}}} \\ $$$$\Rightarrow\:\:\boldsymbol{{tanh}}\mathrm{2}\boldsymbol{{x}}=\frac{\mathrm{2}\frac{\boldsymbol{{shx}}}{\boldsymbol{{chx}}}}{\mathrm{1}+\left(\frac{\boldsymbol{{shx}}}{\boldsymbol{{chx}}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}\boldsymbol{{tanhx}}}{\mathrm{1}+\boldsymbol{{tanh}}^{\mathrm{2}} {x}} \\ $$