Question

已知向量a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),且x属于【π/2，π】，

1）求向量a*b及/a+b/ （2）求函数f(x)=a*b+/a+b/的最大值，并求使函数取得最大值x的值

Answer 1

(1)
a.b
= (cos(3x/2),sin(3x/2)).(cos(x/2),-sin(x/2))
=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos2x
a+b
=(cos(3x/2),sin(3x/2))+(cos(x/2),-sin(x/2))
=(cos(3x/2)+cos(x/2), sin(3x/2)-sin(x/2))
|a+b|^2
=[cos(3x/2)+cos(x/2)]^2+[sin(3x/2)-sin(x/2)]^2
=2 - 2(cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2))
=2-2cos2x
|a+b| =√(2-2cos2x)
(2)
f(x)
=a.b+ |a+b|
=cos2x + √(2-2cos2x)
f'(x) = -2sin2x+ 2sin2x/√(2-2cos2x)
f'(x) =0
sin2x = 0 or
cos2x=1/2 , ( max )
max f(x) = 1/2 + 1 = 3/2