Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Having a problem with an exercise from Gelfand and Saul's Trigonometry, in the section dealing with the half-angle formulae. The exercise (7.a. on p.151) asks the reader to show that:

$$\tan(\alpha/2)\tan(\beta/2)\ +\ \tan(\beta/2)\tan(\gamma/2)\ +\ \tan(\gamma/2)\tan(\alpha/2)\ =\ 1$$

where $\alpha + \beta + \gamma = \pi$.

I made the cotangent substitution for one of the variables and checked it on Wolfram Alpha, which confirms the identity, but for some reason I still can't see how to actually prove it myself. I'm sure I'm missing something very obvious!

share|cite|improve this question
up vote 3 down vote accepted

We have $\dfrac\alpha2 + \dfrac\beta2 + \dfrac\gamma2 =\dfrac\pi2$.


$\cot (\dfrac\alpha2 + \dfrac\beta2 + \dfrac\gamma2) = \large\dfrac{1-\tan\frac\alpha2\tan\frac\beta2 - \tan\frac\beta2\tan\frac\gamma2 - \tan\frac\gamma2\tan\frac\alpha2}{\tan\frac\alpha2+\tan\frac\beta2+\tan\frac\gamma2-\tan\frac\alpha2\tan\frac\beta2\tan\frac\gamma2}=0$ since $\cot\dfrac\pi2=0$.

share|cite|improve this answer
+1 Very clever! I like the three-variable sum formula! – Prism Aug 24 '13 at 13:37

Since $$\tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}$$ we have $$\tan x + \tan y = \tan (x+y) [1-\tan x \tan y].$$ Therefore $$\tan \frac{\alpha}{2}\left( \tan \frac{\beta}{2} + \tan \frac{\gamma}{2} \right) = \tan\left(\frac{\pi}{2} - \frac{\beta}{2}-\frac{\gamma}{2} \right) \tan\left(\frac{\beta}{2}+\frac{\gamma}{2}\right)\left[1-\tan\frac{\beta}{2} \tan\frac{\gamma}{2}\right] = 1-\tan\frac{\beta}{2} \tan\frac{\gamma}{2}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.