Prove $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$ Given $2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$, 
prove that $\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\frac{1}{5}$
Also we know that all the expressions are different from zero and defined. Including the expressions we received during the solution. 
Tried to play with it, didn't seem to work for me.
 A: Applying sum to product rule on LHS and sine of the sum of 2 angles on RHS
$$2\sin(\alpha)+2\sin(\beta)=3\sin(\alpha+\beta)$$
$$4\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha-\beta}{2}\right)=6\sin \left(\frac{\alpha+\beta}{2}\right)\cos \left(\frac{\alpha+\beta}{2}\right)$$
$$2\cos \left(\frac{\alpha}{2}-\frac{\beta}{2}\right)=3\cos \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)$$
$$2\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)+2\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)=3\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)-3\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)$$
$$5\sin \left(\frac{\alpha}{2}\right)\sin \left(\frac{\beta}{2}\right)=\cos\left(\frac{\alpha}{2}\right)\cos \left(\frac{\beta}{2}\right)$$
$$\tan\left(\frac{\alpha}{2}\right)\tan\left(\frac{\beta}{2}\right)=\frac{1}{5} \quad \square$$
A: Recall the angle addition formula $$\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \sin(\beta)\cos(\alpha)$$ for arbitrary angles.
From what you are given,
$$\frac{2}{3}\sin(\alpha)+\frac{2}{3}\sin(\beta)=\sin(\alpha+\beta)$$
So let's say $\cos(\beta) = \frac{2}{3} = cos(\alpha)$
Now your question is reduced to showing that $$ \frac{1}{\sqrt{5}} = \tan\big(\frac{\cos^{-1}(\frac{2}{3})}{2}\big)$$
Option one, reason directly from a triangle.
Option two, you can now use the relevant half-angle formula, 
$$\tan(\frac{\alpha}{2}) = \sqrt{\frac{1-\cos(\alpha)}{1+\cos(\alpha)}}$$
which shows $$\tan(\frac{\alpha}{2}) = \sqrt{\frac{1-2/3}{1+2/3}} = \sqrt{1/5}$$
A: Using the formulas of half-angle substitution, also called Weierstrass substitution formulas,
$$\cos(u)=\dfrac{1-t^2}{1+t^2} , \ \sin(u)=\dfrac{2t}{1+t^2} \ \ \ \text{where} \ \ \ t:=\tan(\tfrac{u}{2}),$$
one has to prove, setting $a=\tan(\alpha/2)$ and $b=\tan(\beta/2)$, that :
$$2\dfrac{2a}{1+a^2}+2\dfrac{2b}{1+b^2}=3 \dfrac{2a}{1+a^2}\dfrac{1-b^2}{1+b^2} +3 \dfrac{2b}{1+b^2}\dfrac{1-a^2}{1+a^2} \implies ab=\dfrac15$$
(we have used $\sin(\alpha+\beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$).
The first expression, once we have multiplied both sides by $(1+a^2)(1+b^2)$ and cancelled by 2, gives:
$$2a+2ab^2+2b+2ba^2 = 3a-3a^2b+3b-3a^2b$$
$$\iff 5ab(a+b)=a+b$$
giving indeed, under the supplementary assumption that $a+b \neq 0$ 

$$ab=\tfrac15.$$

Remark: the overall advantage of formulas (*) is that they convert trigonometrical issues into fully algebraic formulations.
