Find the sum of all the values of $\cos(p\alpha+q\beta)$ , where p and q are positive integers such that $p+q=n$ using the identity Find the sum of all the values of $\cos(p\alpha+q\beta)$ , where p and q are positive integers such that $p+q=n$ using the identity

$\frac{x^{n+1}-y^{n+1}}{x-y}= x^n+x^{n-1}y+x^{n-2}y^{2}+\dots+y^{n}$

I don't see how to proceed with the given information. It is very confusing on my part to understand the question. 
Thanks in advance !
Update: even after getting all the hints. I am not able to proceed to get the desired expression. Kindly share some more.
 A: Hint:
$$
\cos (p\alpha + q \beta) = \mbox{Re}\big((e^{i\alpha})^p \cdot (e^{i\beta})^q \big)
$$
where $q=n-p$
Explanation
\begin{align*}
\sum \limits_{p+q=n}\cos (p\alpha + q \beta) &= \sum \limits_{p+q=n} \mbox{Re}\big((e^{i\alpha})^p \cdot (e^{i\beta})^q \big)\\
\sum \limits_{p=1}^{n-1}\cos (p\alpha + (n-p) \beta) 
&= \mbox{Re} \Big ( \sum \limits_{p=1}^{n-1} (e^{i\alpha})^p \cdot (e^{i\beta})^{n-p} \Big)\quad \mbox{(define $x=e^{i\alpha}$ and $y=e^{i\beta}$)}\\
&= \mbox{Re} \Big ( \sum \limits_{p=1}^{n-1} x^p \cdot y^{n-p} \Big)\\
&= \mbox{Re} \Big ( \sum \limits_{p=0}^{n} x^p \cdot y^{n-p} \Big) - \mbox{Re} \Big (x^n + y^{n} \Big)\\
&= \mbox{Re} \Big ( \frac{x^{n+1} - y^{n+1}}{x-y} \Big) - \mbox{Re} \Big (x^n + y^{n} \Big)\\
&= \mbox{Re} \Big ( xy \frac{x^{n} - y^{n}}{x-y} \Big)\\
&= \mbox{Re} \Big ( e^{i(\alpha + \beta)} \frac{e^{i\alpha n} - e^{i \beta n}}{e^{i \alpha}-e^{i \beta}} \Big)\quad \mbox{(multiply the denom. by conj.)}\\
&= \mbox{Re} \Big ( e^{i(\alpha + \beta)} \frac{(e^{i\alpha n} - e^{i \beta n})(e^{-i \alpha}-e^{-i \beta})}{2 + 2 \cos(\beta- \alpha)} \Big)\\
&= \frac{ \mbox{Re} \Big ( e^{i(\alpha + \beta)}(e^{i\alpha n} - e^{i \beta n})(e^{-i \alpha}-e^{-i \beta}) \Big)}{2 + 2 \cos(\beta- \alpha)} \\
&= \mbox{Exercise  (just distribute the numerator and find the real part)} 
\end{align*}
