f(x)=(cosx+sinx)*(cosx-sinx)+2sinxcosx=cos2x+sin2x=sin(2x+π/4)
(1)T=2π/2=π,
单调递减区间为:2x+π/4属于[2kπ-π/2,2kπ+π/2],解得x属于[kπ-3π/8,kπ+π/4]
(2)当x属于[0,π/2]时,2x+π/4属于[π/4,3π/4],此时,f(x)最大值为1,最小值为2分之根2.
f(x)=(cosx+sinx)*(cosx-sinx)+2sinxcosx=cos2x+sin2x=sin(2x+π/4)
(1)T=2π/2=π,
单调递减区间为:2x+π/4属于[2kπ-π/2,2kπ+π/2],解得x属于[kπ-3π/8,kπ+π/4]
(2)当x属于[0,π/2]时,2x+π/4属于[π/4,3π/4],此时,f(x)最大值为1,最小值为2分之根2.