锐角三角形ABC中,B=π/3,则, 1.sinA - 知识问答

锐角三角形ABC中,B=π/3,则, 1.sinA+cosC 范围为. 2.(cosA)^2+(cosC)^2 范围为?

1个回答

1、sinA+cosC
=sinA+cos(π-A-B)
=sinA-cos(A+π/3)
=sinA-1/2cosA+√3/2sinA
=(1+√3/2)sinA-1/2cosA
=√(2+√3)sin(A-5π/12)
0

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