求值 cos[arcsin(-根号3/2)] 3Q

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求值 cos[arcsin(-根号3/2)] 3Q
求值 cos[arcsin(-根号3/2)] 3Q

求值 cos[arcsin(-根号3/2)] 3Q
arcsin值域是[-π/2,π/2]
sin(-π/3)=-根号3/2
所以原式=cos(-π/3)=cos(π/3)=1/2