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Question Number 82953 by TawaTawa1 last updated on 26/Feb/20

verify that: cosh.cosh^(−1) (y) = y, if y ∈ (1, + ∞)

$$\mathrm{verify}\:\mathrm{that}:\:\:\:\:\:\:\mathrm{cosh}.\mathrm{cosh}^{−\mathrm{1}} \left(\mathrm{y}\right)\:\:\:=\:\:\:\mathrm{y},\:\:\:\:\:\mathrm{if}\:\:\:\:\:\mathrm{y}\:\:\in\:\:\left(\mathrm{1},\:\:\:+\:\infty\right) \\ $$

Commented by mathmax by abdo last updated on 26/Feb/20

we have ch^(−1) (y)=ln(y+(√(y^2 −1))) ⇒ ch(ch^(−1) (y)=ch(ln(y+(√(y^2 −1))) =((e^(ln(y+(√(y^2 −1)))) +e^(−ln(y+(√(y^2 −1)))) )/2) =((y+(√(y^2 −1+))(1/(y+(√(y^2 −1)))))/2) =(((y+(√(y^2 −1)))^2 +1)/(2(y+(√(y^2 −1))))) =((y^2 +2y(√(y^2 −1))+y^2 −1+1)/(2(y+(√(y^2 −1))))) =((2y^2 +2y(√(y^2 −1)))/(2(y+(√(y^2 −1))))) =((y^2 +y(√(y^2 −1)))/(y+(√(y^2 −1))))=((y(y+(√(y^2 −1))))/(y+(√(y^2 −1)))) =y

$${we}\:{have}\:{ch}^{−\mathrm{1}} \left({y}\right)={ln}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow \\ $$$${ch}\left({ch}^{−\mathrm{1}} \left({y}\right)={ch}\left({ln}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)\right.\right. \\ $$$$=\frac{{e}^{{ln}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)} +{e}^{−{ln}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)} }{\mathrm{2}}\:=\frac{{y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}+}\frac{\mathrm{1}}{{y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}}{\mathrm{2}} \\ $$$$=\frac{\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)}\:=\frac{{y}^{\mathrm{2}} \:+\mathrm{2}{y}\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}+{y}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{\mathrm{2}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)} \\ $$$$=\frac{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{y}\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)}\:=\frac{{y}^{\mathrm{2}} \:+{y}\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}{{y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}=\frac{{y}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right)}{{y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}\:={y} \\ $$

Commented by mathmax by abdo last updated on 26/Feb/20

you are welcome

$${you}\:{are}\:{welcome}\: \\ $$

Commented by TawaTawa1 last updated on 26/Feb/20

God bless you sir

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

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