[tex]\begin{gathered} \sin (2\tan ^{-1}(\frac{-\sqrt[\square]{2}}{2})) \\ \\ \text{let }\theta=\tan ^{-1}(\frac{-\sqrt[\square]{2}}{2}) \\ \tan \text{ }\theta=\text{ }\frac{-\sqrt[\square]{2}}{2} \\ \end{gathered}[/tex][tex]\begin{gathered} m^2=2^2+(\sqrt[]{2})^2\text{ (Pythagoras theorem)} \\ m^2=4+2=6 \\ m=\sqrt[]{6} \end{gathered}[/tex][tex]\begin{gathered} -\theta=\text{ }\tan ^{-1}(\frac{-\sqrt[]{2}}{2}) \\ \sin (2\tan ^{-1}(\frac{-\sqrt[\square]{2}}{2}))\text{ becomes sin(-2}\theta) \\ \text{ Applying trig identities, } \\ \sin \text{ (-2}\theta)\text{ = sin(-}\theta-\theta)\text{ = sin(-}\theta)\cos (-\theta)+\cos (-\theta)\sin (-\theta)_{} \\ =\text{ -2sin}\theta\cos \theta \end{gathered}[/tex][tex]\begin{gathered} \text{From the triangle above, } \\ \sin \theta=\frac{\sqrt[]{2}}{\sqrt[]{6}},\text{ cos}\theta=\frac{2}{\sqrt[]{6}} \\ -2\sin \theta\cos \theta=-2\text{ x }\frac{\sqrt[]{2}}{\sqrt[]{6}}\text{ x }\frac{2}{\sqrt[]{6}} \\ \\ =\frac{-4\sqrt[]{2}}{6} \\ \\ =\frac{-2\sqrt[]{2}}{3} \end{gathered}[/tex]
The correct option is the third option
