已知2sin（3π+θ）=cos（π+θ）,求2sin²θ+3sinθcosθ-cos²θ的值.

Question

已知2sin（3π+θ）=cos（π+θ）,求2sin²θ+3sinθcosθ-cos²θ的值.

Answer 1

2sin( 3π+θ) = 2sin( π+θ) = -2sinθ
cos(π+θ) = -cosθ
所以,由2sin( 3π+θ) = cos(π+θ)得：cosθ = 2sinθ
因此,2sin²θ + 3sinθcosθ - cos²θ
= 2sin²θ + 3sinθ*2sinθ - ( 1 - sin²θ)
= 2sin²θ + 6sin²θ - 1 + sin²θ
= 9sin²θ - 1
因为 sin²θ + cos²θ = 1,把cosθ = 2sinθ代入得：
sin²θ + 4sin²θ = 1解得：sin²θ = 1/5
所以,原式 = 9sin²θ - 1 = 9 × 1/5 - 1 = 9/5 - 1 = 4/5