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Question Number 126702 by Eric002 last updated on 30/Dec/20
if tanh(x/2)=t prove that cosh(x)=((1+t^2 )/(1−t^2 ))
$${if}\:\:\:\:{tanh}\frac{{x}}{\mathrm{2}}={t}\:\:{prove}\:{that}\:\:{cosh}\left({x}\right)=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by MJS_new last updated on 23/Dec/20
tanh α =((e^(2α) −1)/(e^(2α) +1))∧cosh α =((e^α +e^(−α) )/2) ⇒ t=tanh (x/2) =((e^x −1)/(e^x +1)) ((1+t^2 )/(1−t^2 ))=((1+(((e^x −1)/(e^x +1)))^2 )/(1−(((e^x −1)/(e^x +1)))^2 ))=((((e^x +1)^2 +(e^x −1)^2 )/((e^x +1)^2 ))/(((e^x +1)^2 −(e^x −1)^2 )/((e^x +1)^2 )))= =((2e^(2x) +2)/(4e^x ))=((e^x +e^(−x) )/2)=cosh x
$$\mathrm{tanh}\:\alpha\:=\frac{\mathrm{e}^{\mathrm{2}\alpha} −\mathrm{1}}{\mathrm{e}^{\mathrm{2}\alpha} +\mathrm{1}}\wedge\mathrm{cosh}\:\alpha\:=\frac{\mathrm{e}^{\alpha} +\mathrm{e}^{−\alpha} }{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${t}=\mathrm{tanh}\:\frac{{x}}{\mathrm{2}}\:=\frac{\mathrm{e}^{{x}} −\mathrm{1}}{\mathrm{e}^{{x}} +\mathrm{1}} \\ $$$$\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}+\left(\frac{\mathrm{e}^{{x}} −\mathrm{1}}{\mathrm{e}^{{x}} +\mathrm{1}}\right)^{\mathrm{2}} }{\mathrm{1}−\left(\frac{\mathrm{e}^{{x}} −\mathrm{1}}{\mathrm{e}^{{x}} +\mathrm{1}}\right)^{\mathrm{2}} }=\frac{\frac{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)^{\mathrm{2}} }}{\frac{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{e}^{{x}} −\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{e}^{{x}} +\mathrm{1}\right)^{\mathrm{2}} }}= \\ $$$$=\frac{\mathrm{2e}^{\mathrm{2}{x}} +\mathrm{2}}{\mathrm{4e}^{{x}} }=\frac{\mathrm{e}^{{x}} +\mathrm{e}^{−{x}} }{\mathrm{2}}=\mathrm{cosh}\:{x} \\ $$

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