Trigonometry in Newtonian Dynamics HARD

A particle of mass $\displaystyle m\sb{1}$ is travelling along the x-axis with velocity $\displaystyle (v\sb{1},0,0)$. It collides elastically with a stationary particle of mass $\displaystyle m\sb{2}$. After the collision, the particle of mass $\displaystyle m\sb{1}$ is travelling with speed $\displaystyle v\sb{1}'$ at angle $\displaystyle a$ to the x-axis, and the particle of mass $\displaystyle m\sb{2}$ is travelling with speed $\displaystyle v\sb{2}'$ at angle $\displaystyle b$ to the x-axis.

The conservation of momentum laws for this collision are:

(i) $\displaystyle m\sb{1}v\sb{1}'sin(a)+m\sb{2}v\sb{2}'sin(b)=0$

(ii) $\displaystyle m\sb{1}v\sb{1}'cos(a)+m\sb{2}v\sb{2}'cos(b)=m\sb{1 }v\sb{1}$

The conservation of energy equation is:

(iii) $\displaystyle m\sb{1}(v\sb{1}')\sp{2}+m\sb{2}(v\sb{2}')\sp{2}=m\ sb{1}(v\sb{1})\sp{2}$

I have to show that these 3 equations imply:

$\displaystyle sin\sp{2}(a+b)=sin\sp{2}(b)+ksin\sp{2}(a)$, where $\displaystyle k=\frac{m\sb{1}}{m\sb{2}}$