sin x = 2/3
opposite side = a = 2
hypotenuse = c= 3
adjacent side = b = ?
b^2 = 9 - 4
b = square root of 5
[tex]\text{ b =- }\sqrt[]{5}[/tex]Second quadrant
sin(2x) = 2sinxcosx
[tex]\begin{gathered} \sin (2x)\text{ = 2(2/3)(}\frac{-\sqrt[]{5}}{3}) \\ \sin (2x)\text{ = -}\frac{4}{3}\cdot\text{ }\frac{\sqrt[]{5}}{3} \\ \sin (2x)\text{ = -}\frac{4\sqrt[]{5}}{9} \end{gathered}[/tex][tex]\begin{gathered} \cos (2x)\text{ = }cos^2x-sin^2x \\ \cos (2x)\text{ = (}\frac{-\sqrt[]{5}}{3})^2\text{ - (}\frac{2}{3})^2 \\ \cos (2x)\text{ = }\frac{5}{9}\text{ - }\frac{4}{3} \\ \cos (2x)\text{ = }\frac{5}{9}\text{ - }\frac{36}{9} \\ \cos (2x)\text{ = }\frac{-31}{9} \end{gathered}[/tex][tex]\text{ tan (2x) = 2}\frac{2}{\sqrt[]{5}}\text{ = }\frac{4}{\sqrt[]{5}}\text{ = }\frac{4\sqrt[]{5}}{5}[/tex]
