Question

已知sina是方程5x2+7x-6=0的根

求[sin(3π/2-a)tan(2π-a)]/[cos(π/2-a)cos(π/2+a)]的值
要详细过程

Answer 1

5x²+7x-6=0
(5x-3)(x+2)=0
x=3/5或x=-2
因为-1≤sina≤1
所以：sina=3/5

[sin(3π/2-a)tan(2π-a)]/[cos(π/2-a)cos(π/2+a)]
=cosatana/(-sin²a)
=-1/sina
=-5/3

Answer 2

[sin(3π/2-a)tan(2π-a)]/[cos(π/2-a)cos(π/2+a)]
=-cosatana/sina(-sina)
=1/sina
因为
sina是方程5x2+7x-6=0
而5x2+7x-6=(5x-3)(x+2)=0
|sina|<=1
所以sina=3/5
即[sin(3π/2-a)tan(2π-a)]/[cos(π/2-a)cos(π/2+a)]=1/(3/5)=5/3