I would like to prove that (1) $$\begin{equation} \tan\left(\frac{\theta}{2}\right)=\tan\left(\frac{\nu}{2}\right)\tan\left(\frac{\pi/2-\epsilon}{2}\right) \end{equation}$$

can transformed to (2) $$x=y+z,$$ where (3) \begin{align} x&=&\mathrm{arctanh}\left(cos(\theta)\right)\\y&=&\mathrm{arctanh}\left(cos(\nu)\right)\\z&=&\mathrm{arctanh}\left(\sin\left(\epsilon\right)\right) \end{align}

By solving for $\theta$ in 1 and 2, we see that these are indeed equal:

enter image description here

For the record, incorrect identity

Initially the question was wrongly stated, and the comments below pertain to this: I would like to prove that (1) $$\begin{equation} \tan\left(\frac{\theta}{2}\right)=\tan\left(\frac{\nu}{2}\right)\tan\left(\frac{\pi/2-\epsilon}{2}\right) \end{equation}$$

can transformed to (2) $$x=y+z,$$ where (3) \begin{align} x&=&\mathrm{arctanh}\left(\theta\right)\\y&=&\mathrm{arctanh}\left(\nu\right)\\z&=&\mathrm{arctanh}\left(\sin\left(\epsilon\right)\right) \end{align}

My attempt on incorrect identity:

Taking the tanh of (2) on both sides and using $\begin{align} \tanh(x + y) &= \frac{\tanh x +\tanh y}{1+ \tanh x \tanh y } \end{align}$ results in (2a) $$\boxed{ \theta = \frac{\nu+\sin\left(\epsilon\right)}{1+\nu \sin\left(\epsilon\right)} }$$

On the other hand, using

\begin{align} \tan \frac{\theta}{2} &= \csc \theta - \cot \theta &= \pm\, \sqrt{1 - \cos \theta \over 1 + \cos \theta} &= \frac{\sin \theta}{1 + \cos \theta} &= \frac{1-\cos \theta}{\sin \theta} \end{align}

and $\sin(\pi/2-\epsilon)=\cos(\epsilon)$ and $\cos(\pi/2-\epsilon)=\sin(\epsilon)$ in (1) yields

$$\boxed{ \begin{equation} \tan\left(\frac{\theta}{2}\right)= \frac{\cos(\epsilon)(1-\cos(\nu))}{\sin(\nu)(1+\sin(\epsilon))} \end{equation} }$$ ... a bit stuck now

  • $\begingroup$ Did you tried starting with tag(y+z) and see if you could derive tag(x) based on the given relation? $\endgroup$
    – Moti
    Dec 24, 2015 at 20:04
  • $\begingroup$ Thank you, but I'm not getting very far. You are saying use tan(arctan) =.... but I dont quite see how this helps now. Otherwise, I know that $\tan(a)\tan(b)=\frac{\cos (A-B)-\cos (A+B)}{\cos (A-B)+\cos (A+B)}$, and that $\operatorname{artanh} \;u \pm \operatorname{artanh} \;v = \operatorname{artanh} \left( \frac{u \pm v}{1 \pm uv} \right)$, but I'm not sure what to follow this up with. $\endgroup$ Dec 27, 2015 at 7:44
  • $\begingroup$ I have a hard time seeing the connection between tan and arctanh here and why it has come up. $\endgroup$ Dec 27, 2015 at 7:47
  • $\begingroup$ I am not saying I know the answer - it requires some work. But start with $\tagh(y+z)$ (I made an error in my first comment), this will lead to a relation between the variables under the arc $\endgroup$
    – Moti
    Dec 27, 2015 at 18:34
  • $\begingroup$ Take a look at my attempt, maybe you have some further suggestions $\endgroup$ Jan 5, 2016 at 2:16

1 Answer 1


Answer to edited question

We'll begin by squaring your equation $(1)$ and invoking appropriate half-angle formulas:

$$\begin{align}\tan^2\frac{\theta}{2}= \tan^2\frac{\nu}{2}\tan^2\left(\frac{\pi}{4}-\frac{\epsilon}{2}\right) \qquad\to\qquad \frac{1-\cos\theta}{1+\cos\theta} &= \frac{1-\cos\nu}{1+\cos\nu}\cdot\frac{1-\cos\left(\frac{\pi}{2}-\epsilon\right)}{1+\cos\left(\frac{\pi}{2}-\epsilon\right)} \\[6pt] &= \frac{1-\cos\nu}{1+\cos\nu}\cdot\frac{1-\sin\epsilon}{1+\sin\epsilon} \end{align}$$ Thus, $$(1-\cos\theta)(1+\cos\nu)(1+\sin\epsilon) = (1+\cos\theta)(1-\cos\nu)(1-\sin\epsilon)$$ so that $$\begin{align} \cos\theta = \frac{\cos\nu + \sin\epsilon}{1+\cos\nu \sin\epsilon} &\quad\to\quad \tanh x = \frac{\tanh y + \tanh z}{1+\tanh y \tanh z}=\tanh(y+z) \\[6pt] &\quad\to\quad x = y+z \quad\square \end{align}$$

Answer to original question

Why do you think this relation is valid?

  • Certainly, if $\epsilon = 0$ (or $z = 0$), then (restricting domains appropriately) $\theta = \nu$ in (1), and $x = y$ in (2), which is consistent.

  • But if, say, $\nu = 0$, then (1) reduces to $\tan(\theta/2) = 0$, so that $\theta = 0$ and $x = 0$. However, $\epsilon$ ---and therefore also $z$--- is arbitrary in (1), whereas (2) requires $z = 0$.

Maybe you have some domain restrictions that avoid the second case, and lots of other numerical examples I could post. Let's look more generally.

Solving for $\theta$ in in terms of $\nu$ and $\epsilon$ in two ways, from (1) and (2)+(3), we find ourselves considering this statement: $$2\;\operatorname{atan}\left(\;\tan\frac{\nu}{2}\;\tan\left(\frac{\pi}{4} - \frac{\epsilon}{2}\right) \;\right) \;\stackrel{\text{?}}{=}\; \tanh\left(\;\operatorname{atanh}\nu + \operatorname{atanh} \sin\epsilon\;\right) \tag{$\star$}$$

Plots of the left- and right-hand sides appear here:

enter image description here

In the image (with $\nu$ and $\epsilon$ bounded by $0$ and $\pi/2$), we see that the plots do match when $\epsilon = 0$ and $\theta = \nu$, as discussed earlier. However, the plots quite clearly pull away from each other, showing a lack of equality for $\epsilon \neq 0$.

Therefore, we "almost-never" satisfy (1) and (2) simultaneously, so that $(\star)$ is not an identity.

  • $\begingroup$ That is true, there is a mistake in the equation that I was given. It doesn't make sense over the domain $\theta \in [0,\pi]$ and $\epsilon \in [0,\pi/2]$ $\endgroup$ Jan 7, 2016 at 4:36
  • $\begingroup$ I will edit my question above to clarify, I found the right relationship. Now, I have to prove it :) $\endgroup$ Jan 7, 2016 at 5:14

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