设0

来源:学生作业帮助网 编辑:作业帮 时间:2024/10/03 17:04:18
设0
x){nMR>_`g$Vwnvr~`{:TS640D9m6f`dA]#-0+uѬ4E1Hu+KyvPـ9ؼ$gdpA]Wh{Ⱦ34jj%^``sn0z F(ᯋp*e.O[Wl^tԧ{Fx] ɜ[sh@<=-F )J%`.E Jz*Z5 c($(` V37uϦl{u; 

设0
设0

设0
sinθ+cosθ=1/5
(sinθ+cosθ)²=(1/5)²
sin²θ+cos²θ+2sinθcosθ=1/25
2sinθcosθ=-24/25
∴(sinθ-cosθ)²=1-2sinθcosθ=1+24/25=49/25
∵0

sinθ+cosθ=1/5
(sinθ+cosθ)^2=(1/5)^2=1/25
(sinθ)^2+2sinθcosθ+(cosθ)^2=1/25
1+2sinθcosθ=1/25
sinθcosθ=-12/25(1)
因0<θ<π,
由(1)得π/2<θ<π(2)
(sinθ-cosθ)^2=(sinθ)^2-2sinθcosθ+(cosθ...

全部展开

sinθ+cosθ=1/5
(sinθ+cosθ)^2=(1/5)^2=1/25
(sinθ)^2+2sinθcosθ+(cosθ)^2=1/25
1+2sinθcosθ=1/25
sinθcosθ=-12/25(1)
因0<θ<π,
由(1)得π/2<θ<π(2)
(sinθ-cosθ)^2=(sinθ)^2-2sinθcosθ+(cosθ)^2=1-2sinθcosθ=49/25
|sinθ-cosθ|=7/5
结合(2)得sinθ-cosθ=7/5(3)
由sinθ+cosθ=1/5 和(3)得

sinθ=4/5
cosθ=-3/5
tanθ =sinθ/cosθ=-4/3

收起