已知向量
(1)当
(2)设x1,x2为函数f(x)=-
|
(1)由
∥a
得:b
cosx+sinx=0,3 2
若cosx=0,则sinx=±1,不合题意.
则tanx=-
.3 2
因此cos2x-sin2x=
=cos2x-2sinxcosx sin2x+cos2x
=1-2tanx tan2x+1
.16 13
(2)f(x)=-
+(2 4
+a
)•b
=(sinx+cosx,b
)•(cosx,-1)-1 2
=(sinx+cosx)cosx-2 4
-1 2
=2 4
sin2x+1 2
cos2x-1 2
=2 4
sin(2x+2 2
)-π 4
.2 4
依题得sin(2x+
)=π 4
,1 2
解得x=k1π-
或x=k2π+π 24
,k1,k2∈Z.7π 24
又|x1-x2|=|k2π+
-k1π+7π 24
|≥π 24
,π 3
所以|x1-x2|的最小值为
.π 3