# Identities with inverse hyperbolic and trigonometric functions, such as $\tanh^{-1} (\cos a)+\tanh^{-1} (\cos b)=\tanh^{-1} (\cos c)$

This was a very surprising discovery for me that identities like this exist:

$$\tan \frac{c}{2}=\tan \frac{a}{2}\tan \frac{b}{2} \qquad \rightarrow$$

$$\tanh^{-1} (\cos c)=\tanh^{-1} (\cos a)+\tanh^{-1} (\cos b)$$

This is a fairly well known one, and can be proven by making substitutions:

$$u=\tan \frac{a}{2}, \qquad v=\tan \frac{b}{2}$$

Another, interesting one exists (proven in the same way):

$$\tan \frac{c}{2}=\frac{\tan \frac{a}{2}-\tan \frac{b}{2}}{\tan \frac{a}{2}+\tan \frac{b}{2}} \qquad \rightarrow$$

$$\tanh^{-1} (\sin c)=\tanh^{-1} (\cos b)-\tanh^{-1} (\cos a)$$

What other identities like this exist?

What is the interpretation of such identities in terms of:

• Complex numbers
• Geometry

Or is it just a coincidense with no particular significance?

Not sure if this is the sort of thing you're after.

$\displaystyle \tanh^{-1}(y) = \frac1{2} \ln \left( \frac{1+y}{1-y} \right)$

Using $\displaystyle \cos(x) = 2\cos^2 \left( \frac{x}{2} \right) - 1$ you get:

$\displaystyle \tanh^{-1}(\cos(x)) = -\ln \left( \tan\left( \frac{x}{2} \right) \right)$

So the $\displaystyle \tanh^{-1}(\cos(x))$ terms have logarithmic properties wrt $\displaystyle \tan \left( \frac{x}{2} \right)$.

This explains the sum of $\displaystyle \tanh^{-1}$ terms producing a product in $\displaystyle \tan$ terms.

It also implies that more terms can be added to the right of the $\displaystyle \tanh^{-1}$ equation and the corresponding $\displaystyle \tan$ terms multiplied to the right of the $\displaystyle \tan$ equation.

• This is exactly what I've been looking for! Thank you! P.S. Since you keep editing, I can't correct your formatting. Please, put \ before each function (such as $\cos$) so they are displayed correctly – Yuriy S Aug 8 '16 at 18:30
• Thanks for the tip. I'm still correcting typos.Cheers – arthur Aug 8 '16 at 18:32
• Well, here is your bounty – Yuriy S Aug 8 '16 at 18:37
• added it to en.wikipedia.org/wiki/… but rewritten but with reference :) – Willemien Nov 3 '16 at 20:48