Evaluating $ 4\sum^{30}_{n=1} n\;T(n)$, where $T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$ 
For $n$ measured in degrees, let 
  $$T(n) = \cos^2(30^\circ -n) - \cos(30^\circ -n)\cos(30^\circ +n) +\cos^2(30^\circ +n)$$
  Evaluate $$ 4\sum^{30}_{n=1} n \cdot T(n)$$

I have tried to use double-angle identities but got stuck with the coefficient $n$. I am new to trig, so I probably miss some advanced concepts. 

Since
$$\cos^2(30^\circ-n)-\cos(30^\circ-n)\cos(30^\circ+n)+\cos^2(30^\circ+n)=\cos(30^\circ-n)[\cos(30^\circ-n)-\cos(30^\circ+n)] +\cos^2(30^\circ+n)=(\frac{\sqrt{3}}{2}\cos n +\frac{1}{2}\sin n )2\sin30°\sin n +(\frac{\sqrt{3}}{2}\cos n -\frac{1}{2}\sin n )^2=
\frac{3}{4}$$
$$ 4\sum^{30}_{n=1} n \cdot T(n)=4\frac{3}{4}(1+2+\cdots+30)=1395 $$
