
1.D【解析】cosa=3/5,a∈(π,2π),∴sina=-4/5
sin(π/4-a)=sin π/4cos a-cosπ/4 sin a=√2/2×3/5-√2/2×(-4/5)=7√2/10故选D
2.B【解析】由a∈(-π/2,π/2),tan a=2√2,得cos a=1/3,所以cos2a=2cos2a-1=2×(1/3)2-1=-7/9故选B
作者:网络 来源: 2022-08-03 11:34:28 阅读:次
1.D【解析】cosa=3/5,a∈(π,2π),∴sina=-4/5 sin(π/4-a)=sin π/4cos a-cosπ/4 sin a=√2/2×3/5-√2/2×(-4/5)=7√2/10故选D 2.B【解析】由a∈(-π/2,π/2),tan a
1.D【解析】cosa=3/5,a∈(π,2π),∴sina=-4/5
sin(π/4-a)=sin π/4cos a-cosπ/4 sin a=√2/2×3/5-√2/2×(-4/5)=7√2/10故选D
2.B【解析】由a∈(-π/2,π/2),tan a=2√2,得cos a=1/3,所以cos2a=2cos2a-1=2×(1/3)2-1=-7/9故选B