Prove the following trigonometric result If $\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$  are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, prove that 
$$\frac{\sin(\theta_1)+ \sin(\theta_2) }{ \cos(\theta_1)+ \cos(\theta_2)} =\cot\phi$$
I have tried with the following process:
Since 
$\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$  are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, we have
 $\sin(\theta_i+\phi)=\frac{1}{2}\sin(2\phi)$, I=1,2.
This gives
  $\sin\theta_i\cos\phi+ \cos\theta_i \sin\phi= \frac{1}{2}\sin(2\phi)$, I=1,2.  
I don't know what will be the next process.

$$\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$$
$$\sin\theta \cos \phi + \sin\phi \cos \theta = \sin \phi \cos\phi$$
$$\sin\theta \cot \phi + \cos \theta = \cos\phi$$
$$\sin\theta \cot \phi - \cos \phi = -\cos\theta$$
Squaring both sides, we get
$$\sin^2\theta \cot^2 \phi + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi= \cos^2\theta=1-\sin^2\theta \tag1$$
$$\sin^2\theta (1+\cot^2 \phi) + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi -1=0$$
$$\sin^2\theta \csc^2 \phi - 2\sin\theta \cot \phi \cos \phi - \sin^2 \phi =0$$
$$\sin^2\theta - 2\sin\theta \sin \phi \cos^2 \phi - \sin^4 \phi =0$$
Hence $\sin \theta_1+\sin \theta_2=2 \sin \phi \cos^2 \phi$
and accordingly, if instead of converting the $\cos$ to $\sin$ in $(1)$, you convert the $\sin$ to $\cos$ , you will get $\cos \theta_1+\cos \theta_2=2 \sin^2 \phi \cos \phi$
The answer will follow. Hope this is clear.
