已知tanθ=2,那么sin²θ+sinθcosθ+cos²θ的值为?

已知tanθ=2,那么sin²θ+sinθcosθ+cos²θ的值为?
已知tanθ=2,那么sin²θ+sinθcosθ+cos²θ的值为?

已知tanθ=2,那么sin²θ+sinθcosθ+cos²θ的值为?
sin²θ+sinθcosθ+cos²θ
=(sin²θ+sinθcosθ+cos²θ)/1
=(sin²θ+sinθcosθ+cos²θ)/(sin²θ+cos²θ)分子分母同时除以cos²θ
=(sin²θ/cos²θ+sinθcosθ/cos²θ+cos²θ/cos²θ)/(sin²θ/cos²θ+cos²θ/cos²θ)
=[(sinθ/cosθ)²+sinθ/cosθ+1]/[(sinθ/cosθ)²+1]
=[tan²θ+tanθ+1]/[tan²θ+1]
=(2²+2+1)/(2²+1)
=(4+3)/(4+1)
=7/5

sin²θ+sinθcosθ+cos²θ
=(sin²θ+sinθcosθ+cos²θ)/1
=(sin²θ+sinθcosθ+cos²θ)/(sin²θ+cos²θ) 注意(1=sin²θ+cos²θ)
分子分母同时除以cos²...

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sin²θ+sinθcosθ+cos²θ
=(sin²θ+sinθcosθ+cos²θ)/1
=(sin²θ+sinθcosθ+cos²θ)/(sin²θ+cos²θ) 注意(1=sin²θ+cos²θ)
分子分母同时除以cos²θ得
(sin²θ/cos²θ+sinθcosθ/cos²θ+cos²θ/cos²θ)/(sin²θ/cos²θ+cos²θ/cos²θ)
=[(sinθ/cosθ)²+sinθ/cosθ+1]/[(sinθ/cosθ)²+1]
=[tan²θ+tanθ+1]/[tan²θ+1]
=(2²+2+1)/(2²+1)
=(4+3)/(4+1)
=7/5

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