求f(x)=2sin(x+π/4)sin(x-π/4)+sin2X的最大值.(答案是根号2,

问题描述:

求f(x)=2sin(x+π/4)sin(x-π/4)+sin2X的最大值.(答案是根号2,

f(x)=2sin(x+π/4)sin(x-π/4)+sin2X=2(根号2/2*sinx+根号2/2*cosx)(根号2/2*sinx-根号2/2*cosx)+sin2x=sinx^2-cosx^2+sin2x=-cos2x+sin2x=根号2sin(2x-π/4)因为sin(2x-π/4)最大值是1所以原式最大值为根号2...您好,请教一下倒数第三步为什么是sin(2x-π/4)而不是sin(2x-1)?谢谢sin2x-cos2x=根号2(根号2/2sin2x-根号2/2cos2x)=根号2(cosπ/4sin2x-sinπ/4cos2x)=根号2sin(2x-π/4)