首先化简
f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx)=[(sinx+cosx)^2+sinx+cosx]/(1+sinx+cosx)=(1+sinx+cosx)(sinx+cosx)/(1+sinx+cosx)=sinx+cosx=√2sin(x+π/4)
故f(x)最小正周期为2π
当x∈[0,2π]时,(x+π/4)∈[π/4,9π/4],则sin(x+π/4)取值范围为[-√2/2,1],则f(x)最小值为-1,最大值为√2.
f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx)（1）求f(x)最小周期（2）求f（x)在[0,2π]上的最值
首先化简
f(x)=(1+sinx+cosx+sin2x)/(1+sinx+cosx)=[(sinx+cosx)^2+sinx+cosx]/(1+sinx+cosx)=(1+sinx+cosx)(sinx+cosx)/(1+sinx+cosx)=sinx+cosx=√2sin(x+π/4)
故f(x)最小正周期为2π
当x∈[0,2π]时,(x+π/4)∈[π/4,9π/4],则sin(x+π/4)取值范围为[-√2/2,1],则f(x)最小值为-1,最大值为√2.