求值:(tan10°-√3)×cos10°/sin50°
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/06/30 21:03:32
求值:(tan10°-√3)×cos10°/sin50°
求证:[sin(2α+β)]/sinα- 2cos(α+β)=sinβ/cosα
化简cosθ+cos(θ+2π/3)+cos(θ+4π/3)
证明:sin(α+β)sin(α-β)=sin^2α-sin^2β
求证:[sin(2α+β)]/sinα- 2cos(α+β)=sinβ/cosα
化简cosθ+cos(θ+2π/3)+cos(θ+4π/3)
证明:sin(α+β)sin(α-β)=sin^2α-sin^2β
(tan10°-√3)×cos10°/sin50°
=(sin10°/cos10°-√3)×cos10°/sin50°
=(sin10°-√3cos10°)/sin50°
=2(cos60°sin10°-sin60°cos10°)/sin50°
=2×sin(10°-60°)/sin50°
=-2
=(sin10°/cos10°-√3)×cos10°/sin50°
=(sin10°-√3cos10°)/sin50°
=2(cos60°sin10°-sin60°cos10°)/sin50°
=2×sin(10°-60°)/sin50°
=-2
求值:(tan10°-√3)×cos10°/sin50°
求值:(tan10°-根号3)cos10°÷sin50°
(tan10°–√3)cos10°/sin50°
化简:cos10°(tan10°-根号3)/sin50°
求值:2sin50°+sin80°(1+3tan10°)1+cos10°
求值:2sin50°+cos10°*(1+根号3tan10°)/cs35°cos40°+cos50°cos55°
化简:(tan10°−3)•cos10°sin50°
sin50°(1+根号3tan10°)求值、
化简求值2sin50°+cos10°(1+3tan10°)cos35°cos40°+cos50°cos55°
(tan10°-根号3)×cos10°÷sin50°
化简(tan10°-根号3)*(cos10°/sin50°)*(2cos10°-sin20°/cos20°)
2sin50°+cos10°(1+3tan10°)1+cos10°