1.
1)sin(θ-π/2)=-sin(π/2-θ)=-cosθ,cos(θ+π/2)=sinθ,所以有cosθ=2sinθ,
所以tanθ=1/2
2)原式=(sinθ+2cosθ)(sinθ-cosθ)=-5sin²θ=-(sin²θ+4sin²θ)=-(sin²θ+cos²θ)=-1
2.由题意有0°<θ<180°,180°
1.
1)sin(θ-π/2)=-sin(π/2-θ)=-cosθ,cos(θ+π/2)=sinθ,所以有cosθ=2sinθ,
所以tanθ=1/2
2)原式=(sinθ+2cosθ)(sinθ-cosθ)=-5sin²θ=-(sin²θ+4sin²θ)=-(sin²θ+cos²θ)=-1
2.由题意有0°<θ<180°,180°