已知:向量a=(sinx,1),b=(cosx,-1/2) 求函数f(x)=a·(a-b)的最大值

问题描述:

已知:向量a=(sinx,1),b=(cosx,-1/2) 求函数f(x)=a·(a-b)的最大值

f(x)
=a.(a-b)
=(sinx,1).(sinx-cosx,3/2)
=(sinx)^2-sinxcosx +3/2
= (1/2)(1-cos2x) -(1/2)sin2x+3/2
=2 -(√2/2)(√2/2)(cos2x+sin2x)
=2- (√2/2)cos(2x-π/4)
max f(x) = 2+√2/2