设有函数f(x)=asin(kx-π/3)和函数g(x)=bcos(2kx-π/6),(a>0,b>0,k>o),若它们的最小正周期之和 为(3π)/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,求这两个函数的解析式.
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![设有函数f(x)=asin(kx-π/3)和函数g(x)=bcos(2kx-π/6),(a>0,b>0,k>o),若它们的最小正周期之和 为(3π)/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,求这两个函数的解析式.](/uploads/image/z/8326696-40-6.jpg?t=%E8%AE%BE%E6%9C%89%E5%87%BD%E6%95%B0f%28x%29%3Dasin%28kx-%CF%80%2F3%29%E5%92%8C%E5%87%BD%E6%95%B0g%28x%EF%BC%89%3Dbcos%282kx-%CF%80%2F6%29%2C%28a%3E0%2Cb%3E0%2Ck%3Eo%29%2C%E8%8B%A5%E5%AE%83%E4%BB%AC%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%E4%B9%8B%E5%92%8C+%E4%B8%BA%EF%BC%883%CF%80%EF%BC%89%2F2%2C%E4%B8%94f%28%CF%80%2F2%29%3Dg%28%CF%80%2F2%29%2Cf%28%CF%80%2F4%29%3D-%E2%88%9A3g%28%CF%80%2F4%29-1%2C%E6%B1%82%E8%BF%99%E4%B8%A4%E4%B8%AA%E5%87%BD%E6%95%B0%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F.)
设有函数f(x)=asin(kx-π/3)和函数g(x)=bcos(2kx-π/6),(a>0,b>0,k>o),若它们的最小正周期之和 为(3π)/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,求这两个函数的解析式.
设有函数f(x)=asin(kx-π/3)和函数g(x)=bcos(2kx-π/6),(a>0,b>0,k>o),若它们的最小正周期之和
为(3π)/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,求这两个函数的解析式.
设有函数f(x)=asin(kx-π/3)和函数g(x)=bcos(2kx-π/6),(a>0,b>0,k>o),若它们的最小正周期之和 为(3π)/2,且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,求这两个函数的解析式.
由它们的最小正周期之和 为(3π)/2
所以2π/k+2π/2k=(3π)/2
解得,k=2〉0
代入原式,得f(x)=asin(2x-π/3)
g(x)=bcos(4x-π/6)
又f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1
代入得,asinπ/3=bsin(-π/6)
asinπ/6=-√3bsinπ/6-1
a>0,b>0
a=1/2,b=√3/2
f(x)=1/2sin(2x-π/3)
g(x)=√3/2cos(4x-π/6)
它们的最小正周期之和 为(3π)/2,那么2π/k+2π/2k=3π,解得k=2,再利用f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1,得asin(π-π/3)=bcos(2π-π/6),
asin(π/2-π/3)=-√3bcos(π-π/6)-1,解得a=1,b=1。