已知tanθ=2求2cosθ-4sinθ/5cosθ+3sinθ和 3sin-2θ-4cos2θ

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/10/11 20:31:07

(2cosθ-4sinθ)/(5cosθ+3sinθ)
=(2cosθ-4cosθ·tanθ)/(5cosθ+3cosθ·tanθ)
=(2-4tanθ)/(5+3tanθ)
=(2-4×2)/(5+3×2)
=-6/11,
3sin(-2θ)-4cos2θ
=-6sinθ·cosθ-4(cos²θ-sin²θ)
=(-6sinθ·cosθ-4cos²θ+4sin²θ)/(sin²θ+cos²θ)
=(-6tanθ·cos²θ-4cos²θ+4tan²θ·cos²θ)/(tan²θ·cos²θ+cos²θ)
=(-6tanθ-4+4tan²θ)/(tan²θ+1)
=(-6×2-4+4×2²)/(2²+1)
=0.
若3sin²θ-4cos²θ
=(3sin²θ-4cos²θ)/(sin²θ+cos²θ)
=(3tan²θ-4)/(tan²θ+1)
=(3×2²-4)/(2²+1)
=8/5
再问：我那个2指的是2次方~
再答：若3sin²θ-4cos2θ =3sin²θ-4(cos²θ-sin²θ) =7sin²θ-4cos²θ =(7sin²θ-4cos²θ)/(sin²θ+cos²θ) =(7tan²θ-4)/(tan²θ+1) =(7×2²-4)/(2²+1) =24/5
再问：你们两个又都弄错了前面那个是3sin-² 不是3sin²
再答： 3sin²(-θ)=3sin²θ
再问：大侠你看清楚点 3sin-²θ =1/3sin²θ
再答：

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