Question

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

Answer 1

(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

追问

我那个2指的是2次方~

追答

若3sin²θ-4cos2θ
=3sin²θ-4(cos²θ-sin²θ)
=7sin²θ-4cos²θ
=(7sin²θ-4cos²θ)/(sin²θ+cos²θ)
=(7tan²θ-4)/(tan²θ+1)
=(7×2²-4)/(2²+1)
=24/5

追问

你们两个又都弄错了 前面 那个是3sin-² 不是3sin²

追答

3sin²(-θ)=3sin²θ

追问

大侠你看清楚点 3sin-²θ =1/3sin²θ

追答

Answer 2

（2cosθ-4sinθ）/（5cosθ+3sinθ）
=[(2cosθ-4sinθ)÷cosθ]/[(5cosθ+3sinθ)÷cosθ]
=[2-4tanθ]/[5+3tanθ]
=[2-4×2]/[5+3×2]
=-6/11
3sin2θ-4cos2θ
=6sinθcosθ-4(2cos²θ-1)
=6sinθcosθ-8cos²θ+4
=(6sinθcosθ-8cos²θ)/(sin²θ+cos²θ)+4
=[(6sinθcosθ-8cos²θ)÷cos²θ]/[(sin²θ+cos²θ)÷cos²θ]+4
=[6tanθ-8]/[tan²θ+1]+4
=[6×2-8]/[2²+2]+4
=4/6+4
=2/3+4
=4又2/3

追问

我那个2是指2次方

追答

3sin²θ-4cos2θ
=3sin²θ-4（1-2sin²θ）
=11sin²θ-4
=11sin²θ/(sin²θ+cos²θ)-4
=[11sin²θ÷cos²θ]/[(sin²θ+cos²θ)÷cos²θ]-4
=[11tan²θ]/[tan²θ++1]-4
=[11×2]/[2²+1]-4
=22/5-4
=4.4-4
=0.4

追问

你们两个又都弄错了 前面 那个是3sin-² 不是3sin²

Answer 3

22