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sin2x Cos2x最小正周期

y=sin2xcos2x=(1/2)sin4x,最小正周期为T=2π/4=π/2.

y=sin2x*cos2x=1/2*2sin2x*cos2x=1/2sin4x 所以最小正周期t=2π/4=π/2

π/2

sin2x+cos2x=√2(sin2x+45°)最小正周期 = 180°

sin2x-cos2x=(根号2)sin(2x θ)(θ是个辅助角,没有关系)所以最小正周期 = 2π÷2=π

y=sin2x-cos2x=2(22sin2x22cos2x)=2sin(2x-π4)∴T=2π2=π故答案为:π

Y=根2(根2/2sin2x-根2/2cos2x) =根2sin(2x-π/4) 因为x的系数为2,所以T=2π/2=π 懂了麻烦采纳,谢谢

sin2x+cos2x=√2sin(2x+π/4)所以最小正周期为2π/2=π

sin2x cos2x=0.5sin4x T=2 π/4=π/2.

cos2x+sin2x=√2sin(2x+π/4)所以最小正周期为T=2π/2=π

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