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/6605800-16-0.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.
T1+T2=2π/|k|+2π/|2k|=3π/k=3π/2
k=2
f(x)=asin(2x-π/3)
g(x)=bcos(4x-π/6)
f(π/2)=asin(2π/3)=asin(π-π/3)=asin(π/3)=acos(π/2-π/6)=acos(π/6)=g(π/2)=bcos(-π/6)=bcos(π/6)
所以a=b
f(π/4)=asin(π/6)=-√3g(π/4)-1=-√3bcos(5π/6)-1
所以a/2=-√3*b(-√3/2)-1=3b/2-1
a=3b-2
a=b
所以a=b=1
f(x)=sin(2x-π/3)
g(x)=cos(4x-π/6)
最小正周期之和 为(3π)/2
则 2π/k+2π/2k=(3π)/2
k=2
f(π/2)=g(π/2),f(π/4)=-√3g(π/4)-1
带入则可以得出答案