速来

问题描述:

速来
f(x)=2sinx(sinx+cosx),求f(x)的最大值

f(x)=2sin²x+2sinxcosx
=2(1-cos2x)/2+sin2x
=√2(sin2x*√2/2-cos2x*√2/2)+1
=√2(sin2xcosπ/4-cos2xsinπ/4)+1
=√2sin(2x-π/4)+1
所以最大值是√2+1