已知向量a=(sinx,3/2), 向量b=(cosx,-1).（1）当向量a//向量b时，求2cos^2x-sin2x的值。
（2）求f(x)＝（向量a+向量b）×向量b在[－π/2,0]上的值域。

1)
向量a//向量b,
sinx: cosx = (3/2):-1
3cosx + 2sinx = 0;
cos^2(x) + sin^2(x) = 1;
sinx = -(3/2)cosx,
cos^2(x) = 4/13;
tanx = -3/2;

2cos^2(x) -sin2x
=2cos^2(x) -2sinx*cosx
=2cos^2(x)*(1-tanx)
=(4/13)*(1+3/2)
=(4/13)*(5/2)
=10/13;

2)
f(x)=（向量a+向量b）×向量b
=(sinx+cosx,1/2)*(cosx,-1)
= (sinx+cosx)*cosx -1/2
= sinxcosx + cos^2(x) -1/2
= sin(2x)/2 + cos(2x)/2
= (sqrt(2)/2)*sin(2x+PI/4);
-PI/2 <= x <=0;
-PI <= 2x <=0;
-3PI/4<=2x+PI/4 <=PI/4;

-1 <= sin(2x+PI/4) <=sqrt(2)/2;

-sqrt(2)/2 <=f(x)<=1/2,

f(x)在[-PI/2,0]的值域是 [-sqrt(2)/2, 1/2].