1b)
Sinx=$\frac{3}{5}$
Ta có : Sin$^{2}$x + cos$^{2}$x=1
⇒Cos$^{2}$x=$\sqrt{1-(\frac{3}{5})}$$^{2}$
⇒Cosx=±$\frac{4}{5}$
do: $\frac{\pi}{2}$ < x < $\pi$
⇒Cosx=-$\frac{4}{5}$
ta có: Sin2x=2Sinx.Cosx= 2.$\frac{3}{5}$.$\frac{-4}{5}$=$\frac{-24}{25}$
Cos2x=2Cos$^{2}$x -1=2.($\frac{3}{5}$)$^{2}$ -1=$\frac{-7}{25}$
