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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 04:23:04
![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](/uploads/image/z/3859815-39-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%881%2F2%2C%E2%88%9A3%2F2%EF%BC%89%2Cb%3D%EF%BC%88cosx%2Csinx%EF%BC%89%2C%EF%BC%881%EF%BC%89a%7C%7Cb%2Cx%E2%88%88%280%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%881%2F2%2C%E2%88%9A3%2F2%EF%BC%89%2Cb%3D%EF%BC%88cosx%2Csinx%EF%BC%89%2C%EF%BC%881%EF%BC%89a%7C%7Cb%2Cx%E2%88%88%280%2C%CF%80%2F2%29%2C%E6%B1%82sinx%E5%92%8Ccos2x%E7%9A%84%E5%80%BC.2.%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3Da%2Ab%2Cf%28a%2B%CF%80%2F3%29%3D12%2F13%2C%E4%B8%94a%E2%88%88%28-%CF%80%2F2%2C0%29%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E5%92%8C)
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
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
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
(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