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求下列函数的最大值,最小值和最小正周期y=sin(x+π/4)+sin(x-π/4)

题目详情
求下列函数的最大值,最小值和最小正周期y=sin(x+π/4)+sin(x-π/4)
▼优质解答
答案和解析
sin(x+π/4)+sin(x-π/4)=sin(x+π/4)+sin[(x+π/4)-π/2]=sin(x+π/4)-sin(x+π/4)cos(π/2)-cos(x+π/4)sin(π/2)=sin(x+π/4)-cos(x+π/4)=根号2[根号2/2sin(x+π/4)-根号2/2cos(x+π/4)]=根号2[sin(x+π/4)cos(π/4)-cos(x+π/4)sin(π/4)]=根号2sin(x+π/4-π/4)=根号2sin(x)
最大值根号2 最小值 -根号2 最小正周期2π