已知向量a=(cos5x/3,sin5x/3),b=(cosx/3,-sinx/3),x∈[0,∏/2]1.求a+b的绝对值2.若f(x)=ab-2λ|a+b|(其中λ大于0)的最小值是-3/2,求λ的值.

问题描述:

已知向量a=(cos5x/3,sin5x/3),b=(cosx/3,-sinx/3),x∈[0,∏/2]
1.求a+b的绝对值
2.若f(x)=ab-2λ|a+b|(其中λ大于0)的最小值是-3/2,求λ的值.

a+b=(cos5x/3+cosx/3,sin5x/3-sinx/3)所以a+b的绝对值=根号下[(cos5x/3+cosx/3)^2+(sin5x/3-sinx/3)^2]=2+2cos5x/3cosx/3-2sin5x/3sinx/3=2+2cos(5x/3+x/3)=2+2cos2x

|a+b|=根号( cos5x/3+ cosx/3,sin5x/3+-sinx/3)
+
= + +2cos5x/cosx/3+ + -2sin5x/3sinx/3
=2+2cos5x/cosx/3-2sin5x/3sinx/3
=2+2(cos5x/cosx/3-2sin5x/3sinx/3)
=2+2cos(5x/3+x/3)
=2+2cos(2x)
a+b的绝对值=根号2+2cos(2x)
2 f(x)=ab-2λ|a+b|=cos2x-2λ(2+2cos(2x))=(4λ+1)cos2x-4λ
当cos2x=0时 f(x)的最小及
f(x)min=(4λ+1)cos2x-4λ=-3/2求得λ=-3/8