已知sinα=5/13,α∈(π/2,π)求sin2α,cos2α,tan2α的值
问题描述：
已知sinα=5/13,α∈(π/2,π)求sin2α,cos2α,tan2α的值
答
sinα=5/13,α∈(π/2,π)
cosα=-12/13(由sinα^2+cosα^2=1,又cosαsin2α=2*sinα*cosα=-60/169
cos2α=2cosα^2-1=119/169
tan2α=sin2α/cos2α=-60/119
答
sina=5/13 α∈(π/2,π) ==>cosa=-12/13
sin2a=2sina*cosa=2*(5/13)*(-12/13)=-120/169
cos2a=1-2sina*sina=1-2*25/169=119/169
tan2a=sin2a/cos2a=-120/119
答
由sinα=5/13,α∈(π/2,π)
则cosα=-12/13(由sinα^2+cosα^2=1,又cosα