Sin(2arccos3/5) вычислить

$arccos \frac{3}{5}=y;\ =\ \textgreater \ cos\ y= \frac{3}{5} ;\ sin(arccos \frac{3}{5})=sin 2y=? \\ \\ sin2y=2sin\ y\cdot cos\ y=2sin\ y \cdot \frac{3}{5}=\frac{6}{5}sin\ y \\ \\ sin^2y=1-cos^2y=1-(\frac{3}{5})^2=1-\frac{9}{25}=\frac{16}{25}\\ sin\ y=\pm \frac{4}{5}$
Т.к. $arccos \frac{3}{5}$ ∈ I четверти, то sin y > 0
$sin(2arccos \frac{3}{5})=\frac{6}{5} \cdot \frac{4}{5}= \frac{24}{25}$