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已知向量a=(1/2,√3/2),b=(cosx,sinx),(1)a||b,x∈(0

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已知向量a=(1/2,√3/2),b=(cosx,sinx),(1)a||b,x∈(0
已知向量a=(1/2,√3/2),b=(cosx,sinx),
(1)a||b,x∈(0,π/2),求sinx和cos2x的值.
2.若函数f(x)=a*b,f(a+π/3)=12/13,且a∈(-π/2,0),求函数f(x)的最小正周期和cos(a-π/3)的值
(1)
二者平行,则其幅角(与+x轴的夹角)相等或相差π向量a幅角为arctan[(√3/2)/(1/2)] = arctan√3 = π/3向量b幅角为π/3或4π/3, 后者超出(0,π/2), 不考虑sinx/cosx = tanx, x = π/3sinx = √3/2cos2x = cos(2π/3) = -1/2(2)f(x) = (1/2)cosx + (√3/2)sinx = sinxcos(π/6) + cosxsin(π/6)= sin(x + π/6)最小正周期为2πα ∈(-π/2,0)f(α + π/3) = sin(α + π/3 + π/6) = sin(α + π/2) = cosα = 12/13sinα = -√(1 - cos²α) = -5/13cos(α - π/3) = cosαcos(π/3) + sinαsin(π/3)= (12/13)(1/2) + (-5/13)(√3/2)= (12 - 5√3)/26