Question

已知sinα是方程5x^2-7x-6=0的根，α是第三象限角

｛[sin(-α-3/2π）cos（3/2π-α）]为分子，cos(π/2-α)sin(π/2+α)]为分母，与tan^2(π-α)相乘，求其值。这个问题我还想问α是第三象限角，是不是加上π就可以将其变成锐角

Answer 1

5x²-7x-6=0
(5x+3)(x-2)=0
则sinα=-3/5
cosα=-4/5
tanα=3/4
π<α<3π/2 -3π/2<-α<-π
∵ 0<3π/2-α<π/2
∴ cos(3/2π-α)=-sinα=3/5
∵ -3π<-α-3π/2<-2π-π/2 即 -π<-α-3π/2<-π/2
∴ sin(-α-3π/2)=cosα=-4/5
∵ -π<π/2-α<-π/2
∴ cos(π/2-α)=sinα=-3/5
∵ 3π/2<π/2+α<2π
∴ sin(π/2+α)=cosα=-4/5
故｛sin(-α-3/2π)cos(3/2π-α)/[cos(π/2-α)sin(π/2+α)]}*tan²(π-α)
={-4/5*3/5/[-3/5*(-4/5)]}*(-3/4)²
=-1*9/16
=-9/16

追问

能不能化简到-tan^2α（老师的答案），麻烦你了， 我先加5分，结束再加一点

追答

π<α<3π/2
-3π/2<-α<-π
-π/2<π-α<0

tan(π-α)=-tanα
tan²(π-α)=tan²α

｛sin(-α-3/2π)cos(3/2π-α)/[cos(π/2-α)sin(π/2+α)]}*tan²(π-α)

={[cosα*(-sinα)]/(sinαcosα)}*tan²α
=-tan²α