Solving for an angle in a Trigonometry equality with Tan functions

I have a problem that says that:

$2 \tan(\beta/2) = \tan(\beta − \alpha)$

Where the value of the $\alpha$ angle is known. What I need to get is the value of $\beta$. How can I obtain it from that?

For context, the figure i'm looking at is this: The $\beta − \alpha$ angle is the $A$ angle from the $ACE$ triangle, and the $\beta/2$ angle is the $A$ angle from the $ADE$ triangle.

Use the formula $$\tan{(x+y)} = \frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$$ on the right hand side to find an expression involving $\tan{\alpha}$ and $\tan{\beta}$, and then you can apply it again to turn the $\tan{\beta}$s into a quadratic expression involving $t:=\tan{(\beta/2)}$. You can then rearrange the equation into a cubic in $t$. This cubic's not very nice, but it might simplify for specific values of $\alpha$.
• Ok, so could you help me with the specifics? I get that: $$2\tan{\beta / 2} = \frac{\tan{\alpha - \tan{\beta}}}{1+\tan{\alpha}\tan{\beta}}$$ So if you let $t =: \beta/2$ Now: $$\tan{\beta} = \tan{t +t} = \frac{\tan{t}+\tan{t}}{1-\tan{t}\tan{t}} = \frac{2\tan{t}}{1 - (\tan{t})^2}$$ So: $$2 \tan{t} = \frac{\tan{\alpha} - \frac{2\tan{t}}{1-(\tan{t})^2}}{1+\tan{\alpha} \frac{2\tan{t}}{1-(\tan{t})^2}}$$ What now? – Cartucho Sep 13 '15 at 17:33
• That looks fine so far, so now you have to rearrange to a polynomial in $\tan{(\beta/2)}$. (You'll find it less writing with $T=\tan{\alpha}$ and $t=\tan{(\beta/2)}$, I suggest). First multiply the fraction on the right by $\frac{1-t^2}{1-t^2}$ to simplify the denominator, then multiply both sides of the equation by the denominator you end up with. The cubic equation I found was $2t^3-3t^2 T-T=0$, which you in general have to solve using the cubic formula, which you can find on Wikipedia. – Chappers Sep 13 '15 at 17:41
• Ahh, I see. I'm getting a different result although. I have that: $$2t = \frac{T -\frac{2t}{1-t^2}}{1+ T\frac{2t}{1-t^2}} \times \frac{1-t^2}{1-t^2}$$ $$2t = \frac{T(1-t^2) - 2t}{(1-t^2)+2Tt}$$ $$2t = \frac{T - Tt^2 -2t}{-t^2 +2Tt +1}$$ $$2t (-t^2 +2Tt +1) = T - Tt^2 -2t$$ $$-2t^3 + 4Tt^2 +2t = T -Tt^2 -2t$$ $$2t^3 -5Tt^2 -4t +T = 0$$ Am I doing something wrong? Thanks a lot for your help – Cartucho Sep 13 '15 at 18:22
• I think you have a sign error in your first formula: you should have $\tan{(\beta-\alpha)} = \frac{\tan{\beta}-\tan{\alpha}}{1+\tan{\beta}\tan{\alpha}}$. Then you'll get the same answer as I did. – Chappers Sep 13 '15 at 18:26