Why don't we take negative values of the argument in $\cosh^{-1}=x\pm \sqrt{x^2+1}$?
and write only
$$\cosh^{-1}x = \ln(x+\sqrt{x^2+1})$$
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why don't you use $\LaTeX$? do you mean $$\cosh^{-1}(x)=\pm\sqrt{x^2+1}$$? – Dr. Sonnhard Graubner Aug 05 '15 at 08:47
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yes @ Dr Sonnhard Graubner it is x+- Sqrt() – Milan Amrut Joshi Aug 05 '15 at 08:48
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Sorry i miss ln in first definition – Milan Amrut Joshi Aug 05 '15 at 08:51
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1To find an inverse function, you have to use a branch of the function that is one to one and onto. With a $+$ sign it is choosing the positive $x$ branch. Think about $\arcsin x$. Its values are only in $[-\pi/2,\pi/2]$ with the same reason. – KittyL Aug 05 '15 at 09:01
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1There should be a minus sign not a plus sign inside the square root. The expression you have written is $\sinh^{-1}x$ – David Quinn Aug 05 '15 at 11:57