First, we'll find sinx using the fundamental trigonometric relation that states that:
sin²x + cos²x = 1
[tex]\begin{gathered} \sin ^2x+(\frac{\sqrt[]{2}}{2})^2=1 \\ \sin ^2x=1-\frac{2}{4} \\ \sin ^2x=\frac{4}{4}-\frac{2}{4} \\ \sin ^2x=\frac{2}{4} \\ \sin x=\pm\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]In the fourth quadrant, cosx is positive, and sinx is negative. So,
[tex]\sin x=-\frac{\sqrt[]{2}}{2}[/tex]We have another trigonometrical relation that states that
[tex]\begin{gathered} \sin 2x=2\cdot\sin x\cdot\cos x \\ \sin 2x=2\cdot-\frac{\sqrt[]{2}}{2}\cdot\frac{\sqrt[]{2}}{2} \\ \sin 2x=\frac{-2\cdot2}{2\cdot2} \\ \sin 2x=-1 \end{gathered}[/tex]Answer: Sin2x = -1
