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
人气:379 ℃ 时间:2019-10-11 14:38:11
解答
{[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
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