(cos40°+sin50°(1+√3tan10°))/sin70°√(1+cos40°)
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/06/18 01:32:45
(cos40°+sin50°(1+√3tan10°))/sin70°√(1+cos40°)
![(cos40°+sin50°(1+√3tan10°))/sin70°√(1+cos40°)](/uploads/image/z/4251521-65-1.jpg?t=%28cos40%C2%B0%2Bsin50%C2%B0%281%2B%E2%88%9A3tan10%C2%B0%29%29%2Fsin70%C2%B0%E2%88%9A%281%2Bcos40%C2%B0%29)
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]
=[cos40°+sin50°×(tan60°-tan10°)/tan50°]/sin70°(√2)cos20°
=[cos40°+(tan60°-tan10°)cos50°]/sin70°(√2)cos20°
=[cos40°+√3cos50°-tan10°cos50°]/ (√2)cos20°^2
=[cos40°+√3sin40°-tan10°sin40°] /√2/2(1+cos40°)
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° /√2/2(1+cos40°)
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°/√2/2(1+cos40°)
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°/√2/2(1+cos40°)
=2cos20°-2cos20°+2(cos20°) ^2/√2/2(1+cos40°)
=1+cos40°/√2/2(1+cos40°)
=√2
=[cos40°+sin50°×(tan60°-tan10°)/tan50°]/sin70°(√2)cos20°
=[cos40°+(tan60°-tan10°)cos50°]/sin70°(√2)cos20°
=[cos40°+√3cos50°-tan10°cos50°]/ (√2)cos20°^2
=[cos40°+√3sin40°-tan10°sin40°] /√2/2(1+cos40°)
=2[(1/2)cos40+(√3/2)sin40°]-(sin10°/cos10°)sin40° /√2/2(1+cos40°)
=2(cos60°cos40°+sin60°sin40°)-[(sin10°)∧2/cos10°sin10°]sin40°/√2/2(1+cos40°)
=2cos20°-[(1-cos20°)/sin20°]2sin20°cos20°/√2/2(1+cos40°)
=2cos20°-2cos20°+2(cos20°) ^2/√2/2(1+cos40°)
=1+cos40°/√2/2(1+cos40°)
=√2
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=
(cos40°+sin50°(1+√3tan10°))/sin70°√(1+cos40°)
[cos40°+sin50°(1+√3tan10°)]/[sin70°√(1+cos40°)]=?
『cos40°+sin50°(1+√3tan10°)』/sin70°√1+cos40°』
求值:cos40°+sin50°(1+3tan10°)sin70°1+sin50°
化简:(cos40+sin50(1+√3tan10))/(sin70√(1+cos40))
[cos40+sin50(1+√3tan10)]/sin70√(1+cos40)
化简sin50°+cos40°(1+√3tan10°)/sin^270°
[cos40+sin50(1+√3tan10)]/sin70√(1+sin50) 希望有具体过程 .
求值{cos40°+sin50°(1+根号3tan10°)}/1+sin50°
求值:cos40°(1+3tan10°)
cos40°+sin50°×(1+√3cot10°)