Trigonometrical Solve There are 2 different values of $ \ \theta \  $. They are $ \ a \ $ and $ \ b \ $, such that $ \ 0 \ < \ a,b \ < \ 360^\circ \ $. 
If $ \ \sin(\theta+\phi) = \frac{1}{2} \sin2\phi \ $ , prove that $$ \ \frac{ \sin a \ + \  \sin b}{\cos a \  + \  \cos b} \ = \ \cot\phi \ . $$

$\sin(a + \phi) = \sin a\cdot \cos\phi + \sin\phi\cdot \cos a = \sin\phi\cdot \cos\phi$
$\sin(b + \phi) = \sin b\cdot \cos\phi + \sin\phi\cdot \cos b = \sin\phi\cdot \cos\phi$
Subtract these equations, and we have:
$(\sin a - \sin b)\cdot \cos\phi - (\cos a - \cos b)\cdot \sin\phi = 0$.
So:
$(\sin a - \sin b)\cdot \cos\phi = (\cos a - \cos b)\cdot \sin\phi$, and we have:
$\dfrac{\sin a - \sin b}{\cos a - \cos b} = \tan\phi$.
Note: please check the right side. It should be $\tan\phi$.
and the sign should be minus.
