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设向量a=(cosx/2,sinx/2)向量b=(sin3x/2,cos3x/2)x∈[0,π/2]

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设向量a=(cosx/2,sinx/2)向量b=(sin3x/2,cos3x/2)x∈[0,π/2]
1.求a,b及a+b的模 2.令函数f(x)=a*b+根号2(a+b)的模,求函数f(x)的最值
设向量a=(cosx/2,sinx/2)向量b=(sin3x/2,cos3x/2)x∈[0,π/2]
(1)
a.b
=(cosx/2,sinx/2).(sin3x/2,cos3x/2)
=sin3x/2cosx/2 + cos3x/2sinx/2
= sin2x
a+b
=(cosx/2,sinx/2)+(sin3x/2,cos3x/2)
= (sin3x/2+cosx/2,cos3x/2+sinx/2)
|a+b|^2
=(sin3x/2+cosx/2)^2+ (cos3x/2+sinx/2)^2
= 2 + 2(sin3x/2cosx/2 + cos3x/2sinx/2)
= 2+ 2sin2x
|a+b| = √(2+2sin2x)
(2)
f(x) = a.b +√2 |a+b|
= sin2x + √2 √(2+2sin2x)
= sin2x + 2√(1+sin2x)
max f(x) when sin2x = 1
max f(x) = 1+ 2√2
min f(x) when sin2x = 0
minf(x) = 2