<s>[INST] Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$ 
I can't seem to wrap my head around that. 
 [/INST] Let $n= \sin x$, so that $n^2 = \sin^{2} x = 1- \cos^{2} x$. Thus
$$
n^{2}- \frac{n^{2}}{2} = 1-\cos^{2} x - \frac{1-\cos^{2}x}{2}.
$$
Multiplying both sides by $2$ yields
$$
2n^{2} - n^{2} = 2-2\cos^{2} x - (1-\cos^{2}x) = 2-2\cos^{2} x - 1 +\cos^{2}x = 1 - \cos^{2} x = \sin^{2} x = n^{2}.
$$
Now dividing both sides by $2$ gives
$$
n^{2} - \frac{n^{2}}{2} = \frac{n^{2}}{2}.
$$
</s>