(1)见解析 (2) f(x) max=5,x=2kπ-
(k∈Z)
(1)证明:假设a与b平行,
则cosxsinx-sinx(cosx+2
)=0,
即sinx=0,与x∈(0,
)时,sinx>0,矛盾.
故a与b不平行.
(2)解:f(x)=a·b-2a·c
=cos 2x+2
cosx+sin 2x-2sinx
=1-2sinx+2
cosx
=1-4sin(x-
).
所以f(x) max=5,x=2kπ-
(k∈Z).
(1)见解析 (2) f(x) max=5,x=2kπ-
(k∈Z)
(1)证明:假设a与b平行,
则cosxsinx-sinx(cosx+2
)=0,
即sinx=0,与x∈(0,
)时,sinx>0,矛盾.
故a与b不平行.
(2)解:f(x)=a·b-2a·c
=cos 2x+2
cosx+sin 2x-2sinx
=1-2sinx+2
cosx
=1-4sin(x-
).
所以f(x) max=5,x=2kπ-
(k∈Z).