Question Number 51887 by Tawa1 last updated on 31/Dec/18
$$\mathrm{Prove}\:\mathrm{that};\:\:\:\:\mathrm{tanh}\left(\mathrm{log}\:\sqrt{\mathrm{3}}\right)\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jan/19
$${e}^{{ix}} ={cosx}+{isinx} \\ $$$${e}^{−{ix}} ={cosx}−{isinx} \\ $$$${x}={i}\theta \\ $$$${e}^{−\theta} ={cos}\left({i}\theta\right)+{isin}\left({i}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:={cosh}\left(\theta\right)−{sinh}\left(\theta\right) \\ $$$${e}^{\theta} ={cosh}\left(\theta\right)+{sinh}\left(\theta\right) \\ $$$${tanh}\left(\theta\right)=\frac{{e}^{\theta} −{e}^{−\theta} }{{e}^{\theta} +{e}^{−\theta} } \\ $$$${tanh}\left({ln}\sqrt{\mathrm{3}}\:\right)=\frac{{e}^{{ln}\sqrt{\mathrm{3}}\:} −{e}^{−{ln}\sqrt{\mathrm{3}\:}} }{{e}^{{ln}\sqrt{\mathrm{3}}\:} +{e}^{−{ln}\sqrt{\mathrm{3}}\:} }=\frac{{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} −{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} }{{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} +{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} }\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} −\left(\mathrm{3}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left(\mathrm{3}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}}.=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{{x}} ={e}^{{xlna}} \\ $$
Commented by Tawa1 last updated on 31/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$