已知tanθ=2,求2sin²θ+3sin2θ+1的值

已知tanθ=2,求2sin²θ+3sin2θ+1的值
已知tanθ=2,求2sin²θ+3sin2θ+1的值

已知tanθ=2,求2sin²θ+3sin2θ+1的值
tanθ=sinθ/cosθ=2
sinθ=2cosθ
sin²θ+cos²θ=1
(2cosθ)²+cos²θ=1
5cos²θ=1
cos²θ=1/5
2sin²θ+3sin2θ+1
=2(2cosθ)²+3×(2sinθcosθ)+1
=8cos²θ+12cos²θ+1
=20cos²θ+1
=20×(1/5)+1
=4+1
=5

5
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