Question

已知sinα是方程5x²-7x-6=0的根，求值过程

求[sin(-α-3/2π)sin(3/2π-α)tan²(2π-α)]/[cos(π/2-α)cos(π/2+α)cos²(π-α)]的值
过程!!

Answer 1

由5x²-7x-6=0 得 (5x+3)(x-2)=0
sina=-3/5或sina=2（舍去）下面是每一步化简过程
sin(-α-3/2π)=-cosa sin(3/2π-α)=cosa tan²(2π-α)=tan²α
cos(π/2-α)=sina cos(π/2+α)=-sina cos²(π-α)]=cos²α
最后代入 得 原式=25/16

Answer 2

5x²-7x-6=0
(5x+3)(x-2)=0
sina=-3/5或sina=2舍去
[sin(-α-3/2π)sin(3/2π-α)tan²(2π-α)]/[cos(π/2-α)cos(π/2+α)cos²(π-α)]
=[sin(π/2-a)sin(a-π/2)tan²a]/[-sinasinacos²a]
=-cos²atan²a/(-sin²acos²a)
=1/cos²a
=1/(1-sin²a)
=1/(1-9/25)
=25/16