设arccos(-1/4)=α,α属于〔0,π〕则cosα=-1/4,sinα=√〔1-(-1/4)²〕=√15/4
sin[兀/3 + arc cos(-1/4)]
=sin(π/3+α)
=sinπ/3cosα+cosπ/3sinα
=√3/2*(-1/4)+1/2*√15/4
=-√3/8+√15/8
=(√15-√3)/8
设arccos(-1/4)=α,α属于〔0,π〕则cosα=-1/4,sinα=√〔1-(-1/4)²〕=√15/4
sin[兀/3 + arc cos(-1/4)]
=sin(π/3+α)
=sinπ/3cosα+cosπ/3sinα
=√3/2*(-1/4)+1/2*√15/4
=-√3/8+√15/8
=(√15-√3)/8