题目

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

提问时间：2020-10-10

答案

｛[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