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Question Number 178357 by Spillover last updated on 15/Oct/22

Given that sinh^(−1) x=sech^(−1) x show x=(√((((√5)−1)/2) ))

$$\mathrm{Given}\:\mathrm{that}\:\mathrm{sinh}\:^{−\mathrm{1}} \mathrm{x}=\mathrm{sech}\:^{−\mathrm{1}} \mathrm{x}\:\:\: \\ $$$$\mathrm{show} \\ $$$$\mathrm{x}=\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:} \\ $$

Answered by Ar Brandon last updated on 16/Oct/22

sinh^(−1) x=sech^(−1) x ϕ=sinh^(−1) x ⇒sinhϕ=x cosh^2 ϕ−sinh^2 ϕ=1 ⇒sechϕ=(1/( (√(1+x^2 )))) ⇒ϕ=sech^(−1) ((1/( (√(1+x^2 ))))) ⇒sech^(−1) ((1/( (√(1+x^2 )))))=sech^(−1) x ⇒(1/( (√(1+x^2 ))))=x ⇒1=x^2 (x^2 +1) , x>0 ⇒x^4 +x^2 −1=0 ⇒x^2 =((−1+(√5))/2) ⇒x=(√(((√5)−1)/2))

$$\mathrm{sinh}^{−\mathrm{1}} {x}=\mathrm{sech}^{−\mathrm{1}} {x} \\ $$$$\varphi=\mathrm{sinh}^{−\mathrm{1}} {x}\:\Rightarrow\mathrm{sinh}\varphi={x} \\ $$$$\mathrm{cosh}^{\mathrm{2}} \varphi−\mathrm{sinh}^{\mathrm{2}} \varphi=\mathrm{1}\:\Rightarrow\mathrm{sech}\varphi=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow\varphi=\mathrm{sech}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow\mathrm{sech}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)=\mathrm{sech}^{−\mathrm{1}} {x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}={x}\:\Rightarrow\mathrm{1}={x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\:,\:{x}>\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}} \\ $$

Commented by Spillover last updated on 16/Oct/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

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