已知向量a=(sinx,1),b=(cosx,-1/2),求函数f(x)=a(a-b)的最小正周期,及当0
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已知向量a=(sinx,1),b=(cosx,-1/2),求函数f(x)=a(a-b)的最小正周期,及当0
已知向量a=(sinx,1),b=(cosx,-1/2),求函数f(x)=a(a-b)的最小正周期,及当0<=x<=π/2时f(x)的最大最小值
已知向量a=(sinx,1),b=(cosx,-1/2),求函数f(x)=a(a-b)的最小正周期,及当0
f(x)=(sinx,1)(sinx-cosx,3/2) f(x)max=5/2 , f(x)min=2-根号2/2
=(sinx)^2-sinxcosx+3/2
=(1-cos2x)/2-1/2sin2x+3/2
=-1/2(cos2x+sin2x)+2
=-根号2/2sin(2x+π/4)+2
因为 0<=x<=π/2
所以 π/4<=(2x+π/4)<=5π/4