求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期

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求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期y=si

求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期
求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期

求y=sin(2x+π/4)+cos(2x-π/4)的最小正周期
y=sin(2x+π/4)+cos(2x-π/4)
=sin(2x+π/4)+cos[(2x+π/4)-π/2]
=sin(2x+π/4)+sin(2x+π/4)
=2sin(2x+π/4)
∴ 最小正周期T=2π/2=π