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