Question

已知sinα是方程5x^2-7x-6=0的根α是第三象限，

已知sinα是方程5x^2-7x-6=0的根α是第三象限，则｛[sin(α-3/2π）cos（3/2π-α）]/[cos(π/2-α)sin(π/2+α)]｝×tan^2(π-α)=
｛[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

Answer 1

｛[sin(α-3/2π）cos（3/2π-α）]/[cos(π/2-α)sin(π/2+α)]｝×tan^2(π-α)
=｛-sin(3/2π-a)cos（3/2π-α）]/[cos(π/2-α)sin(π/2+α)]｝×tan^2(π-α)
=｛-1/2sin(3π-2a)]/[cos(π/2-α)sin(π/2+α)]｝×tan²α
=｛-1/2sin2a]/[sinacosa]｝tan²α
=｛-sin2a]/[2sinacosa]｝tan²α
=-tan²α
5x^2-7x-6=0
1 -2
5 3
(x-2)(5x+3)=0
sina=-3/5
cosa=-4/5
tana=3/4
tan²a=9/16
∴-9/16

Answer 2

5x^2-7x-6=0
(5x+3)(x-2)=0
x=-3/5或x=2
根据题意
sina=-3/5或sina=2（舍去）
｛[sin(α-3/2π）cos（3/2π-α）]/[cos(π/2-α)sin(π/2+α)]｝×tan^2(π-α)=
={[sin(α+π/2)cos(π/a+a)]/[sinacosa]}*tan²a
=[sin(2a+π)/sin2a]*tan²a
=[-sin2a/sin2a]*tan²a
=-tan²a
sina=-3/5
cosa=-4/5
tana=3/4
原式=-9/16

Answer 3

下面那个式子容易理解错，麻烦把式子解释一下。我巳解决a的值。