# A "known" tangent half-angle formula?

In another posting I wrote about a trigonometric relation I had derived, but that ended up not being the main point of the posting:

Strange pattern in rounding errors?

So as long as we're here, let's make it the main point of this posting. I posted something like this a couple of days ago on stackexchange, with no answers yet.

Let's make this two questions:

• Is this "known" in the sense of being in some book, refereed paper, or the like?;
• Is there a straightforward way to prove this?

Here's how I derived this relation: I showed that a certain function satisfies a certain differential equation; then I showed that a certain function emerges as the antiderivative that you get by the usual second-semester-calculus methods; then I said these ought to be the same thing because they both solve a geometry problem that arose in some amateur cartography of a (maybe?) somewhat impractical sort, therefore they must be the same; then I checked it numerically and it checked. But there ought to be a more straightforward way.

Here's the result: If $$\tan\gamma=\dfrac{\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$$ and $\alpha, \beta, \gamma\in(0,\pi)$ or $\alpha,\beta,\gamma\in(-\pi,0)$ then $$\tan\dfrac\gamma2=\tan\dfrac\alpha2\cdot\tan\dfrac\beta2.$$

A tangent half-angle formula that everyone knows, or at least that's out there in trigonometry-for-adults books that were occasionally published before about 1930, says $$\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta} = \tan\frac{\alpha+\beta}{2}.$$ Does it make any sort of sense to say that the fact that what I derived, and this "known" identity are reminiscent of each other has some reason behind it? (So I guess this is really three questions.)

• I arrived at this identity in a horribly roundabout way, and as expected, several people have shown that it reduces to high-school trigonometry exercises, so at least in the proof, this proposition might be considered "trivial". But I think maybe in some of the consequences it might not be. But I don't feel like being specific about that just yet. Commented Dec 14, 2012 at 17:02
• Do you feel like being specific now? Commented Jul 1, 2014 at 13:52
• @GerryMyerson : It actually came from thinking about Villarceau circles. Commented Jul 2, 2014 at 14:50

## 3 Answers

Using the tangent double-angle formula $\tan\gamma=\frac{2\tan\tfrac{\gamma}{2}}{1-\tan^2\tfrac{\gamma}{2}}$ we get \begin{align} \tan\gamma & = \frac{2\tan\tfrac{\beta}{2}\tan\tfrac{\alpha}{2}}{1-\tan^2\tfrac{\beta}{2}\tan^2\tfrac{\alpha}{2}} \\[10pt] & = \frac{2\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2}\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}}{\cos^2\tfrac{\beta}{2}\cos^2\tfrac{\alpha}{2}-\sin^2\tfrac{\beta}{2}\sin^2\tfrac{\alpha}{2}} \\[10pt] & =\frac{\sin\beta\sin\alpha}{2(\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}-\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2})(\cos\tfrac{\beta}{2}\cos\tfrac{\alpha}{2}+\sin\tfrac{\beta}{2}\sin\tfrac{\alpha}{2})} \\[10pt] & =\frac{\sin\beta\sin\alpha}{2\cos\tfrac{\beta+\alpha}{2}\cos\tfrac{\beta-\alpha}{2}} \\[10pt] & =\frac{\sin\beta\sin\alpha}{\cos\beta+\cos\alpha} \end{align}

• Technically, this is proving $A$ implies $B$, where the question asked for a proof that $B$ implies $A$. But I suppose all the steps are reversible. Commented Dec 13, 2012 at 22:52
• @GerryMyerson : The one thing that's not quite reversible is this: You can't just let $\gamma=\arctan\dfrac{\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$; rather, you have to choose the approrpriate one of two points on the circle where the tangent has a given value. Although $\tan\gamma$ is the same regardless of which of those you pick, $\tan(\gamma/2)$ is not. That question need not be mentioned if you do the proof in the direction seen in this answer, but for the converse of that, the issue comes up. Commented Dec 14, 2012 at 16:47
• I probably ought to have said "if and only if" in my question, rather than just going one direction. Commented Dec 14, 2012 at 16:47
• @Gerry: You are right. This is a proof that $\tan\gamma=\ldots$ if $\tan(\gamma/2)=\ldots$, rather than the converse. Assuming $\tan(\gamma)=\ldots$, we get merely that $f(\tan(\gamma/2))=f(\tan(\beta/2)\tan(\alpha/2))$, where $f(t)=2t/(1-t^2)$, as you note in your answer. Commented Dec 14, 2012 at 18:05

Another way: let $r=\tan\alpha/2$, $s=\tan\beta/2$, $t=\tan\gamma/2$. Then $\sin\alpha=2r/(1+r^2)$, $\cos\alpha=(1-r^2)/(1+r^2)$, $\tan\gamma=2t/(1-t^2)$, and $${\sin\alpha\sin\beta\over\cos\alpha+\cos\beta}$$ reduces to $2rs/(1-(rs)^2)$, so the question reduces to deriving $rs=t$ from $${2t\over1-t^2}={2rs\over1-(rs)^2}$$

Here's one way to derive the identity:

Suppose $\tan(\gamma)=\frac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)+\cos(\beta)}$. Multiplying this equation through by $\cos(\gamma)$ gives an expression for $\sin(\gamma)$ in terms of $\cos(\gamma)$ and functions of $\alpha$, $\beta$. We now take the Pythagorean identity $\cos^2(\gamma)+\sin^2(\gamma)=1$, and replace $\sin(\gamma)$ with the expression derived above, getting $$\left(\frac{1+\cos(\alpha)\cos(\beta)}{\cos(\alpha)+\cos(\beta)}\cos(\gamma)\right)^2=1$$ Some simplification was done, but the only trig identity used was the Pythagorean identity.

This yields $\cos(\gamma)=\pm\frac{\cos(\alpha)+\cos(\beta)}{1+\cos(\alpha)\cos(\beta)}$, $\sin(\gamma)=\pm\frac{\sin(\alpha)\sin(\beta)}{1+\cos(\alpha)\cos(\beta)}$. We assume $\alpha,\beta,\gamma\in(0,\pi)$, so this forces the $\pm$ signs to be $+$ (it seems like this identity fails when $\alpha,\beta,\gamma<0$).

We have a tangent half-angle formula $\tan(\gamma/2)=\frac{\sin(\gamma)}{1+\cos(\gamma)}$. Combining with the formulas for $\sin(\gamma)$, $\cos(\gamma)$ gives $$\tan(\gamma/2)=\frac{\sin(\alpha)\sin(\beta)}{(1+\cos(\alpha))(1+\cos(\beta))}$$ Using the tangent half-angle formula again gives the desired identity.

• But the identity does work when $\alpha,\beta,\gamma<0$. Commented Dec 14, 2012 at 16:56