Sin(arccos√3/3+arcsin(-1/2)) помогитеееее

sin(arccos√3/3 +arcsin(-1/2))  ; ****************************************************************************** sin(α +β) =sinα*cosβ +cosα*sinβ   [α = arccos√3/3 ;β =arcsin(-1/2) ] ****************************************************************************** sin(arccos√3/3 +arcsin(-1/2))= sin(arccos√3/3)*cos(arcsin(-1/2)) + cos(arccos√3/3)*sin(arcsin(-1/2))  = sin(arcsin√6/3)*cos(arcccos(√3/2)+√3/3*(-1/2) =√6/3*√3/2 - √3)/6 =(3√2 -√3)/6 .