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.)

  • $\begingroup$ 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. $\endgroup$ Dec 14, 2012 at 17:02
  • 3
    $\begingroup$ Do you feel like being specific now? $\endgroup$ Jul 1, 2014 at 13:52
  • $\begingroup$ @GerryMyerson : It actually came from thinking about Villarceau circles. $\endgroup$ Jul 2, 2014 at 14:50

3 Answers 3


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

  • $\begingroup$ 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. $\endgroup$ Dec 13, 2012 at 22:52
  • $\begingroup$ @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. $\endgroup$ Dec 14, 2012 at 16:47
  • $\begingroup$ I probably ought to have said "if and only if" in my question, rather than just going one direction. $\endgroup$ Dec 14, 2012 at 16:47
  • $\begingroup$ @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. $\endgroup$ 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.

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

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