化简:[sin50°+cos40°(1+根号3tan10°)]/sin²70°
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:2024/07/08 04:57:32
化简:[sin50°+cos40°(1+根号3tan10°)]/sin²70°
![化简:[sin50°+cos40°(1+根号3tan10°)]/sin²70°](/uploads/image/z/1365416-8-6.jpg?t=%E5%8C%96%E7%AE%80%EF%BC%9A%5Bsin50%C2%B0%2Bcos40%C2%B0%281%2B%E6%A0%B9%E5%8F%B73tan10%C2%B0%29%5D%2Fsin%26sup2%3B70%C2%B0)
cos40度(1+根号3tan10度)
=2(cos40/cos10)((1/2)cos10+((根号3)/2)sin10)
=2(cos40/cos10)sin(10+30)
=sin80/cos10
=1
所以:
[sin50°+cos40°(1+根号3tan10°)]/sin²70°
=[sin50度+cos40度(1+根号3tan10度)]/cos²20°
=(1+sin50)/(cos20)^2
=(1+sin50)/[(1+cos40)/2]
=2(1+sin50)/(1+cos40)
=2(1+cos40)/(1+cos40)
=2
=2(cos40/cos10)((1/2)cos10+((根号3)/2)sin10)
=2(cos40/cos10)sin(10+30)
=sin80/cos10
=1
所以:
[sin50°+cos40°(1+根号3tan10°)]/sin²70°
=[sin50度+cos40度(1+根号3tan10度)]/cos²20°
=(1+sin50)/(cos20)^2
=(1+sin50)/[(1+cos40)/2]
=2(1+sin50)/(1+cos40)
=2(1+cos40)/(1+cos40)
=2
化简:[sin50°+cos40°(1+根号3tan10°)]/sin²70°
化简{sin50°+cos40°(1+根号3*tan10°)}/sin平方70°,详细步骤
化简sin50°+cos40°(1+√3tan10°)/sin^270°
求值{cos40°+sin50°(1+根号3tan10°)}/1+sin50°
[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°
【2sin50°+sin10°(1+根号3tan10°】*根号2sin²80°=
sin50°(1+根号3tan10°)
化简:[2sin50°+sin10°(1+根号3*tan10°)]*根号sin^280°
求值:2sin50°+cos10°*(1+根号3tan10°)/cs35°cos40°+cos50°cos55°