As [tex]157.5^\circ[/tex] is in the second quadrant, [tex]\sin {157.5}^\circ>0[/tex].
Hence,
[tex]\sin{157.5}^\circ=+\sqrt{1-\cos^2{157.5}^\circ}=\sqrt{1-\dfrac{1+\cos{\left(157.5\cdot 2\right)}^\circ}{2}}=\medskip\\=\sqrt{\dfrac{2-1-\cos{315}^\circ}{2}}=\sqrt{\dfrac{1-\cos\left({360}^\circ-{45}^\circ\right)}{2}}=\sqrt{\dfrac{1-\cos{45}^\circ}{2}}=\medskip\\=\sqrt{\dfrac{1-\frac{\sqrt{2}}{2}}{2}}=\sqrt{\dfrac{2-\sqrt{2}}{4}}=\dfrac{\sqrt{2-\sqrt{2}}}{2}[/tex]
Which is answer A.
