Найти ответ sin(3p/2+arccos(1/3)) = ? sin(2arccos(3/5)) = ?

$sin( \frac{3 \pi }{2}+arccos \frac{1}{3} )= \\ = sin(\frac{3 \pi }{2})*cos(arccos \frac{1}{3})+cos(\frac{3 \pi }{2})*sin(arccos \frac{1}{3})= \\ =- \frac{1}{3} \\ sin(2arccos \frac{3}{5} )=2*sin(arccos \frac{3}{5})*cos(arccos \frac{3}{5})= \\ =2* \sqrt{1- \frac{9}{25} } * \frac{3}{5} = \frac{6}{5} * \frac{4}{5} = \frac{24}{25} =0.96$
cos(arccos(a))=a
sin(arccos(a))=√(1-a²)
cos(arcsin(a))=√(1-a²)
sin(2a)=2sin(a)cos(a)
sin(a+b)=sin(a)*cos(b)+cos(a)sin(b).