关于三角的化简1.5π<α<2π,要求化简 [根号(1-cosα)+根号(1+cosα)]÷[根号(1-cosα)-根号
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/01 21:44:34
关于三角的化简
1.5π<α<2π,要求化简 [根号(1-cosα)+根号(1+cosα)]÷[根号(1-cosα)-根号(1+cosα)]+根号(1+sinα)÷根号(1-sinα)
1.5π<α<2π,要求化简 [根号(1-cosα)+根号(1+cosα)]÷[根号(1-cosα)-根号(1+cosα)]+根号(1+sinα)÷根号(1-sinα)
[√(1-cosα)+√(1+cosα)]÷[√(1-cosα)-√(1+cosα)]+√(1+sinα)÷√(1-sinα)
=[sinα/2-cosα/2]/[sinα/2+cosα/2) + (sinα/2+cosα/2)/(sinα/2-cosα/2)
=2[(sinα/2)^2+(cosα/2)^2]/[(sinα/2)^2-(cosα/2)^2]
=2/(-cosα)
√(1-cosα)=√2(sinα/2)
√(1+cosα)=-√2(cosα/2) cosα/2
=[sinα/2-cosα/2]/[sinα/2+cosα/2) + (sinα/2+cosα/2)/(sinα/2-cosα/2)
=2[(sinα/2)^2+(cosα/2)^2]/[(sinα/2)^2-(cosα/2)^2]
=2/(-cosα)
√(1-cosα)=√2(sinα/2)
√(1+cosα)=-√2(cosα/2) cosα/2
化简(1+sinα/根号1+cosα-根号1-cosα)+(1-sinα/根号1+cosα+根号1-cosα)
若π<α<2π,化简根号1-cosα
化简根号下(1-2sinα/2·cosα/2) + 根号下(1+2sinα/2·cosα/2)
化简 根号((1-cosα)/2)
根号(1-2×sinα×cosα)化简
化简 :根号下(1-2sinαcosα)+根号下(1+2sinαcosα)
化简根号下[(1-cos x)/(1+cos x)] + 根号下[(1+cos x)/(1-cos x)]
已知sin(3π-α)=根号二cos(3π/2+β),根号三cos(-α)=负根号二cos(π+β),且0
已知sin(5π-α)=根号2×cos[(2÷7)+β]和根号3×cos(-α)=根号2×cos(π+α),且0<α<π
已知cos(π\3-2)=1\3,则cosα+根号3sinα=
1.根号下[(1-cosα)/(1+cosα)]+根号下[(1+cosα)/(1-cosα)](α是钝角)
化简根号下[(1-cos x)/(1+cos x)] + 根号下[(1+cos x)/(1-cos x)]其中x属于(π