# Inverse hyperbolic function expressed as inverse trigonometry function

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## Main Question or Discussion Point

Consider $y=\cos{-x}=\cos x=\cosh ix$.

Thus, $\pm x=\cos^{-1}y$ and $ix=\cosh^{-1}y$.

So $\cosh^{-1}y=\pm i\cos^{-1}y$.

Renaming the variable $y$, we have $\cosh^{-1}x=\pm i\cos^{-1}x$.

Next, we evaluate the derivative of $\cosh^{-1}x$ by converting it to $\cos^{-1}x$ using $\cosh^{-1}x=i\cos^{-1}x$.

$\frac{d}{dx}\cosh^{-1}x=\frac{d}{dx}i\cos^{-1}x$
$=i\frac{-1}{\sqrt{1-x^2}}$ --- (*)
$=\frac{-i}{i\sqrt{x^2-1}}$ where I've used the rule $\sqrt{-c}=i\sqrt{c}$ for $c>0$, since $x>1$.
$=\frac{-1}{\sqrt{x^2-1}}$, which differs from the correct answer by a negative sign.

This means we would get the correct answer had we used $\cosh^{-1}x=-i\cos^{-1}x$.

My question is how do we know when to add the negative sign.

I realized that step (*) is wrong because the formula for the derivative of $\cos^{-1}$ is true only for $-1<x<1$. However, since we get the right answer in the end, it seems like there is a reason why we could still use the formula for values of $x$ outside $-1<x<1$, and an explanation when we should use $\cosh^{-1}x=-i\cos^{-1}x$ instead of $\cosh^{-1}x=i\cos^{-1}x$.

Buzz Bloom
Gold Member
Hi Happiness:

I am not sure I understand what you are asking, but it seems to me that the first equation omitted
cosh ix = cosh -ix.​
That is, both cos and cosh are symmetrical functions. Therefore when you take the inverse, you get two possible correct answers:
+ and -. This the same as taking the square root of a positive value and getting both a positive and negative result.

Hope this helps.

Regards,
Buzz

Hi Buzz

I am asking why is $\cosh^{-1}x=-i\cos^{-1}x$? And why isn't $\cosh^{-1}x=+i\cos^{-1}x$? Why is the positive result rejected?

Buzz Bloom
Gold Member
Why is the positive result rejected?
Hi Happiness:

Can you post a link to where you found that the positive result is rejected?

Regards,
Buzz

Buzz Bloom
Gold Member
Hi Happiness:

I replied to this in the thread

Regards,
Buzz