<s>[INST] If $\cos A+\cos B=p$ and $\sin A+\sin B=q$, then find $\cos\left( \frac {A+B}{2}\right)$ in terms of $p$ and $q$ If $\cos A+\cos B=p$ and $\sin A+\sin B=q$ then find $\cos \left( \dfrac {A+B}{2}\right)$ in terms of $p$ and $q$. 
My Attempt:
$$\cos A+\cos B=p$$
$$2\cos \left( \dfrac {A+B}{2}\right)\cos \left( \dfrac {A-B}{2}\right)=p$$
And,
$$\sin A+ \sin B=q$$
$$2\sin \left( \dfrac {A+B}{2} \right)\cos \left( \dfrac {A-B}{2} \right)=q$$
Now,
$$\tan \left( \dfrac {A+B}{2}\right)=\dfrac {q}{p}$$.
How do I proceed further?
 [/INST] here is a geometric interpretation of your work: think of the points $(\cos A, \sin A)$ and $(\cos B , \sin B)$ as two points also called $A$ and $B$ on the unit circle. the arc length is measured from the initial point $O = (1,0).$ now,
the mid point of the chord $AB$ has the coordinate $(p/2, q/2).$ we push this point onto the midpoint of the arc $(A+B)/2$ by dividing by its length to get the cordinates  $$\left(\frac p{\sqrt{p^2 + q^2}}, \frac q{\sqrt{p^2+q^2}}\right).$$
the $x$-coordinate $\frac p{\sqrt{p^2 + q^2}}$ is $\cos((A+B)/2).$
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