设0
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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...
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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
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