速来f（x）=2sinx（sinx+cosx),求f（x）的最大值
问题描述：
速来
f（x）=2sinx（sinx+cosx),求f（x）的最大值
答
求导
f`(x)=4sinxcosx+2cos2x
=2(sin2x+cos2x)
=2根号2sin(2x+pi/4)
答
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