有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,
问题描述:
有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为3π/2,
且f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,求a、b、k的值
答
f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),已知它们的周期和为2π/k+π/k=3π/2,∴k=2.又f(π/2)=g(π/2),f(π/4)=-√3g(π/4)+1,∴-a(√3)/2=-b√3,a/2=-√3*b(1-√3)/(1+√3)+1,化简得...