What is the value of $\sin^2(6°)- \sin^2(12°)+ \sin^2(18)-\cdots \text{till $15$th term}$? Lately a friend of mine asked me above question.   
After reaching this 
$$-\cos(12)+\cos(24)-\cos(36)+\cos(48)-\cos(60)+\cos(72)-\cos(84)+ 1, $$
I could not simply it any further but later I noticed that there is a general pattern here
$L=\sin^2(x)- \sin^2(2x)+ \sin^2 (3x)-\cdots\pm\sin^2 (kx)$,  
where $kx=90^\circ$ and $k \in \mathbb{R} $

$L = \cfrac{1}{2} \text{ for $k$ odd}$ 
$L= -\cfrac{1}{2} \text{ for $k$ even}$  
And $M=\cos^2(x)- \cos^2(2x)+ \cos^2 (3x) - \cdots \pm \cos^2 (kx)$
$M= \dfrac{1}{2}$ for $k=1,2,3,\ldots$
Can anyone explain it to me?
 A: We have that $$\sum_{k=1}^{15}(-1)^k\sin^2(6°k)=\frac{1}{2}\sum_{k=1}^{15}(-1)^k(1-\cos(12°k))=-\frac{1}{2}+S=-\frac{1}{2}$$
where $S=0$ because
$$S=\sum_{k=1}^{15}(-1)^k \cos(12°k)=\mbox{Re}\left(\sum_{k=1}^{15}(-1)^ke^{ik\pi/15}\right)\\=\mbox{Re}\left(-e^{i\pi/15}\cdot \frac{1+e^{i15\pi/15}}{1+e^{i\pi/15}}\right)=\mbox{Re}(0)=0.$$ 
A: Hint: You can rewrite the first equation because ${sin}^2(kx)=\dfrac{1}{2}(1-cos(2kx))$:
$$\sum_{k=0}^n{(-1)^n\sin^2(kx)}=\sum_{k=0}^n(-1)^n{\dfrac{1}{2}}(1-cos(2kx))$$
$$\sum_{k=0}^n(-1)^n{\dfrac{1}{2}}(1-cos(2kx))=\sum_{k=0}^n(-1)^n{\dfrac{1}{2}}-\dfrac{1}{2}\sum_{k=0}^n(-1)^ncos(2kx)$$
A: If $S=\cos12^\circ-\cos24^\circ+\cos36^\circ-\cdots$ upto $n$ terms
As $\cos(180^\circ+y)=-\cos y,\cos(360^\circ+y)=+\cos y$
$$-S=\sum_{r=1}^n\cos\{r(180+12)^\circ\}$$
Using $\sum \cos$ when angles are in arithmetic progression,
$$-S=\dfrac{\sin\left(n\cdot\dfrac{192^\circ}2\right)\cos\left(192^\circ+\dfrac{(n-1)192^\circ}2\right)}{\sin\dfrac{192^\circ}2}=\dfrac{\sin(n\cdot96^\circ)\cos(n+1)96^\circ}{\sin96^\circ}$$
Here $n=15$
