10.导学号93774101已知向量a=(cos x,sin x),b=(-cos x,cos x),c=(-1,0).

(1)若x=π/6,求向量a,c的夹角;

(2)当x∈[π/2 "," 9π/8]时,求函数f(x)=2a·b+1的最大值.

解(1)∵a=(cos x,sin x),c=(-1,0),

　　∴|a|=√(cos^2 x+sin^2 x)=1,|c|=√("(-" 1")" ^2+0^2 )=1.

　　当x=π/6时,a=(cos π/6 "," sin π/6)=(√3/2 "," 1/2),

　　a·c=√3/2×(-1)+1/2×0=-√3/2,cos=(a"·" c)/("|" a"|·|" c"|" )=-√3/2.∵0≤≤π,∴=5π/6.

　　(2)f(x)=2a·b+1=2(-cos2x+sin xcos x)+1

　　=2sin xcos x-(2cos2x-1)=sin 2x-cos 2x

　　=√2sin(2x"-" π/4).

　　∵x∈[π/2 "," 9π/8],∴2x-π/4∈[3π/4 "," 2π],

　　∴sin(2x"-" π/4)∈["-" 1"," √2/2],

　　∴当2x-π/4=3π/4,即x=π/2时,f(x)max=1.