Question

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

Answer 1

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