Question

已知tan(α-π)=-1/2,求下列各式的值.（要详细过程）

已知tan(θ-π)=-1/2,求下列各式的值.（1）sin²θ-2sin(π-θ)×sin(θ+2/π)-cos²(π-θ)4sin²(θ-π/2)-3cos²(θ+3π/2) （2）sin²(θ+kπ)+3cos(3π/2-θ+kπ)×sin(3π/2+θ+kπ)-1

Answer 1

tan(θ-π)=-1/2
-tan(π-θ)=-1/2
tanθ = -1/2
(1)
sin²θ-2sin(π-θ)×sin(θ+π/2)-cos²(π-θ)4sin²(θ-π/2)-3cos²(θ+3π/2)
=sin²θ-2sinθ×cosθ-cos²θ(4cos²θ)-3sin²θ
=-2sin²θ-2sinθcosθ-4(cosθ)^4
=-2(1/5)+ 2(2/5)-4(2/5)^2
=(-10+20-8)/25
=-2/25

(2)
sin²(θ+kπ)+3cos(3π/2-θ+kπ)×sin(3π/2+θ+kπ)-1
=sin²θ+3sinθ×cosθ-1
= 1/5 - 3(2/5) +1
=(1-6+5)/5
=0

追问

我的答案上第一题是94/25,而第二题是-2啊

追答

(1)
sin²θ-2sin(π-θ)×sin(θ+π/2)-cos²(π-θ)4sin²(θ-π/2)-3cos²(θ+3π/2)
我想清楚
cos²(π-θ)4sin²(θ-π/2)
是
cos²[(π-θ)4].sin²(θ-π/2)
还是
cos²(π-θ)x(4sin²(θ-π/2))
(2)
sin²(θ+kπ)+3cos(3π/2-θ+kπ)×sin(3π/2+θ+kπ)-1
=sin²θ-3sinθ×cosθ-1
= 1/5 - 3(2/5) -1
=(1-6-5)/5
=-2

追问

cos²(π-θ)x(4sin²(θ-π/2))

追答

(1)
sin²θ-2sin(π-θ)×sin(θ+π/2)-cos²(π-θ)4sin²(θ-π/2)-3cos²(θ+3π/2)
=sin²θ-2sinθ×cosθ-cos²θ(4cos²θ)-3sin²θ
=-2sin²θ-2sinθcosθ-4(cosθ)^4
=-2(1/5)+ 2(2/5)-4(4/5)^2
=(-10+20-64)/25
=-54/25
我己经反复查过
如果
sin²θ+2sin(π-θ)×sin(θ+π/2)-cos²(π-θ)4sin²(θ-π/2)-3cos²(θ+3π/2)
=-2sin²θ+2sinθcosθ-4(cosθ)^4
=-2(1/5)- 2(2/5)-4(4/5)^2
=(-10-20-64)/25
=-94/25