已知向量a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),且x属于【π/2,π】,
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
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
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(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
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
1年前
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