向量a=(cosx,sinx),向量b(-cosx,cosx),向量c(-1,0)
(1)
x=π/6
a=(√3/2,1/2)
|a|=1 |c|=1
cos=a*c/(|a|*|c)=-√3/2
所以 =150°
(2)
x∈[π/2,9π/8]
f(x)=2ab+1
=2(cosx,sinx)(-cosx,cosx)+1
=2(-cos²x+sinxcosx)+1
=-2cos²x+2sinxcosx+1
=-cos2x+sin2x
=√2sin(2x-π/4)
因为 x∈[π/2,9π/8]
所以 2x-π/4∈[3π/4,2π]
得 sin(2x-π/4)∈[-1,√2/2]
则 f(x)∈[-√2,1]
则 最大值为1