已知ABC是锐角三角形ABC的三个内角,且向量a=(tanA,-sinA)b=(1/2sin2A,cosB)向量a,b的夹角为α(1)求证:0≤α<π/2(2)求函数f(α)=2sin^2(π/4+α)- 根号三cos2α的最大值
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 17:22:09
![已知ABC是锐角三角形ABC的三个内角,且向量a=(tanA,-sinA)b=(1/2sin2A,cosB)向量a,b的夹角为α(1)求证:0≤α<π/2(2)求函数f(α)=2sin^2(π/4+α)- 根号三cos2α的最大值](/uploads/image/z/8881147-19-7.jpg?t=%E5%B7%B2%E7%9F%A5ABC%E6%98%AF%E9%94%90%E8%A7%92%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E4%B8%AA%E5%86%85%E8%A7%92%2C%E4%B8%94%E5%90%91%E9%87%8Fa%3D%28tanA%2C-sinA%29b%3D%281%2F2sin2A%2CcosB%29%E5%90%91%E9%87%8Fa%2Cb%E7%9A%84%E5%A4%B9%E8%A7%92%E4%B8%BA%CE%B1%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A0%E2%89%A4%CE%B1%EF%BC%9C%CF%80%2F2%EF%BC%882%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%28%CE%B1%29%3D2sin%5E2%28%CF%80%2F4%2B%CE%B1%EF%BC%89-+%E6%A0%B9%E5%8F%B7%E4%B8%89cos2%CE%B1%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
已知ABC是锐角三角形ABC的三个内角,且向量a=(tanA,-sinA)b=(1/2sin2A,cosB)向量a,b的夹角为α(1)求证:0≤α<π/2(2)求函数f(α)=2sin^2(π/4+α)- 根号三cos2α的最大值
已知ABC是锐角三角形ABC的三个内角,且向量a=(tanA,-sinA)b=(1/2sin2A,cosB)向量a,b的夹角为α
(1)求证:0≤α<π/2
(2)求函数f(α)=2sin^2(π/4+α)- 根号三cos2α的最大值
已知ABC是锐角三角形ABC的三个内角,且向量a=(tanA,-sinA)b=(1/2sin2A,cosB)向量a,b的夹角为α(1)求证:0≤α<π/2(2)求函数f(α)=2sin^2(π/4+α)- 根号三cos2α的最大值
(1)a=(tanA,-sinA),b=(1/2sin2A,cosB)
a●b=tanA*1/2sin2A-sinAcosB
=sinA/cosA*sinAcosA-sinAcosB
=sin A-sinAcosB
=sinA(1-cosB)
∴sinA(1-cosB)>0
又tanA*cosB+sinA*1/2sin2A
=sinAcosB/cosA+sin AcosA
∵A,B,C是锐角三角形的三个内角
sinAcosB/cosA+sin AcosA>0
∴向量a,b不共线
∴向量a,b夹角范围是(0,π/2)
(2)f(x)=2sin^2(π/4+x)- 根号三cos2x
=1-cos(π/2+2x)-根号三cos2x
=sin2x-根号三2x+1
=2sin(2x-π/3)+1
∵ x∈[0,π/2)
∴2x-π/3∈[-π/3,2π/3)
∴f(x)的最大值为3
所以用x代替了.累死我了.