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I've recently started learning hyperbolic functions and inverse hyperbolic functions, and I came across this equation involving inverse hyperbolic functions. I tried to solve it numerically (I got x=-0.747), but how would you solve it analytically? I don't know how to type in latex so please forgive me.

$$\sinh^{-1}(x) + \cosh^{-1}(x+2) = 0$$

metamorphy
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Kola
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    Have you considered arithmetic manipulations, such as moving one term or the other to the opposite side of the equation, taking $\sinh$ or $\cosh$ of both sides, etc.? Where did you encounter the listed problem? – abiessu Jul 11 '22 at 00:07
  • $-0.747$ seems awfully close to negative simple fraction, no? – Oscar Lanzi Jul 11 '22 at 01:21

2 Answers2

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We have

$$\sinh^{-1}(x)=-\cosh^{-1}(x+2)$$

Apply $\cosh^2(x)$ on both sides

$$\cosh^2(\sinh^{-1}(x))=\cosh^2(-\cosh^{-1}(x+2))=(x+2)^2$$

since $\cosh$ is even. Since $\cosh^2(x)=1+\sinh^2(x)$ we have on the LHS

$$\cosh^2(\sinh^{-1}(x))=1+\sinh^2(\sinh^{-1}(x))=1+x^2$$

So $$1+x^2=(x+2)^2$$ $$ 1+x^2=x^2+4x+4$$ $$-3=4x$$ and finally $$x=-\frac{3}{4}$$

jjagmath
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  • Thank you very much, jjagmath. I tried to cosh both sides of the equation but didn't get very far so proceed to do it numerically. Now I realise I should have cosh^2 both sides of the equation. – Kola Jul 11 '22 at 10:04
  • @Abiessu, I saw this problem online on CIMT hyperbolic functions (google that) click the pdf link. The question is on Exercise 2D, question 4. – Kola Jul 11 '22 at 10:15
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HINTS:

Drawing circular and hyperbolic function triangles for inverse functions may be helpful:

enter image description here

Identities

$$ \cosh^2 x - \sinh ^2 x = 1 $$

$$ \cosh (a +b)= \cosh a \cosh b + \sinh a \sinh b $$

Given equation

$$ \cosh^{-1}\sqrt{1+x^2} + \cosh^{-1}(x+2) =0 $$

Take cosh both sides

$$\sqrt{1+x^2}(x+2) + ..... =1 $$

Can you take it forward?

Another easier way:

$$ \sinh^{-1 }(-x) = \cosh^{-1}(x+2)$$

$$ \cosh^{-1}\sqrt{1+x^2}= \cosh^{-1}(x+2) \to $$

Take cosh both sides and simplify algebra.

$$ 1+x^2 = x^2+4 x +4 ; 4x+3=0 \;; x=-\frac34. $$

Narasimham
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