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.
$$\sinh^{-1}(x) + \cosh^{-1}(x+2) = 0$$
On
HINTS:
Drawing circular and hyperbolic function triangles for inverse functions may be helpful:
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.
$$ 1+x^2 = x^2+4 x +4 ; 4x+3=0 \;; x=-\frac34. $$
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 $$x=-\frac{3}{4}$$