y=sin(π/3 - 2x)+cos2x 的最小正周期是?
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y=sin(π/3 - 2x)+cos2x 的最小正周期是?
y = sin(π/3 - 2x) + cos2x
= sin(π/3)cos2x - cos(π/3)sin2x + cos2x
= (sin(π/3) + 1)cos2x - cos(π/3)sin2x
= acos2x - bsin2x (a,b为两个常数)
= c * ( a/c*cos2x - b/c*sin2x )(c = (a+b)^0.5)
= c*sin(2x + d)
T = 2π/2 = π
= sin(π/3)cos2x - cos(π/3)sin2x + cos2x
= (sin(π/3) + 1)cos2x - cos(π/3)sin2x
= acos2x - bsin2x (a,b为两个常数)
= c * ( a/c*cos2x - b/c*sin2x )(c = (a+b)^0.5)
= c*sin(2x + d)
T = 2π/2 = π
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