sin theta + sin phi = a

_{cos theta + cos phi = b}

_{theta- phi = alpha}

_{eliminate theta and phi}

sin(theta) +sin(phi)=a implies 2sin((theta +phi)/2)cos((theta-phi)/2)=a “equation1” similarly 2cos((theta+phi)/2)cos((theta-phi)/2)=b “equation 2” on dividing both equations we have tan((theta+phi)/2)=a/b implies sin((theta+phi)/2)=a/(a²+b²)^1/2 substituting in equation  one we have cos(alpha/2)=(a²+b²)^1/2/2