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 $$ \ \frac{ \sin a \ + \ \sin b}{\cos a \ + \ \cos b} \ = \ \cot\phi \ . $$
On
$\sin(a + \phi) = \sin a\cdot \cos\phi + \sin\phi\cdot \cos a = \sin\phi\cdot \cos\phi$
$\sin(b + \phi) = \sin b\cdot \cos\phi + \sin\phi\cdot \cos b = \sin\phi\cdot \cos\phi$
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.
We have $\displaystyle\sin\theta\cos\phi=\sin\phi(\cos\phi-\cos\theta)$
Squaring we get $\displaystyle\sin^2\theta\cos^2\phi=\sin^2\phi(\cos\phi-\cos\theta)^2$
$\displaystyle\implies \sin^2\phi(\cos^2\phi+\cos^2\theta-2\cos\phi\cos\theta)=(1-\cos^2\theta)\cos^2\phi$
$\displaystyle\iff \cos^2\theta-2\cos\phi\sin^2\phi\cos\theta+\sin^2\phi\cos^2\phi-\cos^2\phi=0$
whose roots are $\displaystyle\cos a,\cos b\implies \cos a+\cos b=2\cos\phi\sin^2\phi$
Similarly starting with, $\displaystyle\cos\theta\sin\phi=\cos\phi(\sin\phi-\sin\theta),$
we shall find $\displaystyle\sin a+\sin b=2\cos^2\phi\sin\phi$