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 $$\cos^2(30^\circ-n)-\cos(30^\circ-n)\cos(30^\circ+n)+\cos^2(30^\circ+n)=\cos(30^\circ-n)[\cos(30^\circ-n)-\cos(30^\circ+n)] +\cos^2(30^\circ+n)=(\frac{\sqrt{3}}{2}\cos n +\frac{1}{2}\sin n )2\sin30°\sin n +(\frac{\sqrt{3}}{2}\cos n -\frac{1}{2}\sin n )^2= \frac{3}{4}$$
$$ 4\sum^{30}_{n=1} n \cdot T(n)=4\frac{3}{4}(1+2+\cdots+30)=1395 $$
Like Ginger bread,
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\cos^2x+\cos^2y=1+\cos(x-y)\cos(x+y)$$
$$\cos x\cos y=\dfrac{\cos(x-y)+\cos(x+y)}2$$
If $\cos(x+y)=\dfrac12\iff x+y=360^\circ m\pm60^\circ,$
$$T(n)=1-\dfrac{\dfrac12}2=?$$
Here $x=30^\circ-n,y=30^\circ+n$
Well it uses a neat identity that cos(A+B)*cos(A-B)= cos^2A-sin^2B
just expand cos(30-n) = cos30cos(n) +sin30sin(n). and similarly cos(30+n) for middle term
cos(30-n)cos(30+n) = cos^2(30) - sin^2(n)
now simplify to get T(n) = 3/4 hence ans is sum= 3*30*31/2 = 1395.
