[tex]\sin^2\dfrac{7\pi}8=\dfrac{1-\cos\dfrac{7\pi}4}2=\dfrac{1-\dfrac1{\sqrt2}}2=\dfrac{\sqrt2-1}{2\sqrt2}[/tex]
Since [tex]\dfrac{7\pi}8[/tex] lies in the interval [tex]0<x<\pi[/tex], and [tex]\sin x>0[/tex] in this interval, you know that when you take the square root, you should consider the positive root only.
So,
[tex]\sin\dfrac{7\pi}8=\sqrt{\dfrac{\sqrt2-1}{2\sqrt2}}[/tex]
