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Question Number 51887 by Tawa1 last updated on 31/Dec/18

Prove that; tanh(log (√3)) = (1/2)

$$\mathrm{Prove}\:\mathrm{that};\:\:\:\:\mathrm{tanh}\left(\mathrm{log}\:\sqrt{\mathrm{3}}\right)\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jan/19

e^(ix) =cosx+isinx e^(−ix) =cosx−isinx x=iθ e^(−θ) =cos(iθ)+isin(iθ) =cosh(θ)−sinh(θ) e^θ =cosh(θ)+sinh(θ) tanh(θ)=((e^θ −e^(−θ) )/(e^θ +e^(−θ) )) tanh(ln(√3) )=((e^(ln(√3) ) −e^(−ln(√(3 ))) )/(e^(ln(√3) ) +e^(−ln(√3) ) ))=((e^((1/2)ln3) −e^(−(1/2)ln3) )/(e^((1/2)ln3) +e^(−(1/2)ln3) )) . =(((3)^(1/2) −(3)^((−1)/2) )/((3)^(1/2) +(3)^((−1)/2) ))=((3^((1/2)+(1/2)) −1)/(3^((1/2)+(1/2)) +1)).=((3−1)/(3+1))=(1/2) a^x =e^(xlna)

$${e}^{{ix}} ={cosx}+{isinx} \\ $$$${e}^{−{ix}} ={cosx}−{isinx} \\ $$$${x}={i}\theta \\ $$$${e}^{−\theta} ={cos}\left({i}\theta\right)+{isin}\left({i}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:={cosh}\left(\theta\right)−{sinh}\left(\theta\right) \\ $$$${e}^{\theta} ={cosh}\left(\theta\right)+{sinh}\left(\theta\right) \\ $$$${tanh}\left(\theta\right)=\frac{{e}^{\theta} −{e}^{−\theta} }{{e}^{\theta} +{e}^{−\theta} } \\ $$$${tanh}\left({ln}\sqrt{\mathrm{3}}\:\right)=\frac{{e}^{{ln}\sqrt{\mathrm{3}}\:} −{e}^{−{ln}\sqrt{\mathrm{3}\:}} }{{e}^{{ln}\sqrt{\mathrm{3}}\:} +{e}^{−{ln}\sqrt{\mathrm{3}}\:} }=\frac{{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} −{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} }{{e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} +{e}^{−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}} }\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} −\left(\mathrm{3}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{3}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} +\left(\mathrm{3}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}}{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}}.=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{{x}} ={e}^{{xlna}} \\ $$

Commented by Tawa1 last updated on 31/Dec/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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