Paper 1 · November 2010 · hyperbolic functions

Given that $y = \sinh^{-1} x$ , express $y$ in logarithmic form.

Model answer

ln(x+sqrt(x^2+1))

Also accepted: ln(x + sqrt(x^2 + 1)), ln(x+√(x²+1))

Explanation

Since

x=\sinh y=\tfrac12\left(e^y-e^{-y}\right)

e^{2y}-2xe^y-1=0

. Solving as a quadratic in

e^y

e^y=x\pm\sqrt{x^2+1}

. The negative root is less than zero and

e^y>0

always, so

e^y=x+\sqrt{x^2+1}

and

y=\ln\left(x+\sqrt{x^2+1}\right)

. The same steps with

2x

in place of

x

give

\sinh^{-1}2x=\ln\left(2x+\sqrt{4x^2+1}\right)

Derived from ZIMSEC Further Mathematics 9187/1, November 2010, Q2

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