[tex]\it sin2x = 2sinxcosx
\\ \\
x\in\left(0,\ \dfrac{\pi}{2}\right) \Rightarrow cosx=\sqrt{1-sin^2x} =\sqrt{1-\left(\dfrac{4}{5}\right)^2}=\sqrt{1-\dfrac{16}{25}}=
\\ \\ \\
= \sqrt{\dfrac{9}{25}}=\dfrac{3}{5}
\\ \\ \\
sin2x=2\cdot\dfrac{4}{5}\cdot\dfrac{3}{5} = \dfrac{24}{25}[/tex]
