来自梅兴的问题
求sin(A+10°)·[1-√3tan(A-10°)]的值,其中A=60°
求sin(A+10°)·[1-√3tan(A-10°)]的值,其中A=60°
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2020-01-26 22:22
求sin(A+10°)·[1-√3tan(A-10°)]的值,其中A=60°
求sin(A+10°)·[1-√3tan(A-10°)]的值,其中A=60°
sin(A+10°)*【1-根号3tan(A-10°)】=sin(60°+10°)*[1-√3tan(60°-10°)]=sin70°*[1-√3tan50°=2sin70°*[(1/2)cos50°-(√3/2)sin50°]/cos50°=2sin70°sin(30°-50°)/sin40°=-2sin20°cos20°/sin40°...