若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =
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若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =
![若sinθ=(m-3)/(m+5),cosθ=(4-2m)/(m+5),θ∈(π/2,π),则m= ,tanθ =](/uploads/image/z/6016242-66-2.jpg?t=%E8%8B%A5sin%CE%B8%3D%28m-3%29%2F%28m%2B5%29%2Ccos%CE%B8%3D%284-2m%29%2F%28m%2B5%29%2C%CE%B8%E2%88%88%28%CF%80%2F2%2C%CF%80%29%2C%E5%88%99m%3D+%2Ctan%CE%B8+%3D)
sin²θ+cos²θ=1
[(m-3)/(m+5)]²+[(4-2m)/(m+5)]²=1
(m-3)²+(4-2m)²=(m+5)²
m²-6m+9+16-16m+4m²=m²+10m+25
4m²-32m-0
m=0或m=8
(1) m=0 ,sinθ=-3/5 与θ∈(π/2, π), 矛盾
(2)m=8
siinθ=5/13,cosθ=-12/13
所以 m=8,tan θ=sinθ/cosθ= -5/12
再问: sinθ=-3/5 与θ∈(π/2, π),为什么矛盾
再答: θ∈(π/2, π), sinθ>0 所以矛盾
[(m-3)/(m+5)]²+[(4-2m)/(m+5)]²=1
(m-3)²+(4-2m)²=(m+5)²
m²-6m+9+16-16m+4m²=m²+10m+25
4m²-32m-0
m=0或m=8
(1) m=0 ,sinθ=-3/5 与θ∈(π/2, π), 矛盾
(2)m=8
siinθ=5/13,cosθ=-12/13
所以 m=8,tan θ=sinθ/cosθ= -5/12
再问: sinθ=-3/5 与θ∈(π/2, π),为什么矛盾
再答: θ∈(π/2, π), sinθ>0 所以矛盾
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