已知函数F(x)=根号[(1-t)/(1+t)],g(x)=cosx*f(sinx)+sinx*f(cosx)(1)将函数g(x)化简成Asin(wx+a)+B(A>0,w>0,a∈〔0,2π〕)的形式(2)求g(x)的值域a∈【0,2π】x∈[0,17π/12]f(x)改为f(t)

问题描述:

  已知函数F(x)=根号[(1-t)/(1+t)],g(x)=cosx*f(sinx)+sinx*f(cosx)

  (1)将函数g(x)化简成Asin(wx+a)+B(A>0,w>0,a∈〔0,2π〕)的形式

  (2)求g(x)的值域

  a∈【0,2π】x∈[0,17π/12]f(x)改为f(t)

  (1)f(t)=√[(1-t)/(1+t)]   f(sinx)=√[(1-sinx)/(1+sinx)]   f(cosx)=√[(1-cosx)/(1+cosx)]   g(x)=cosx*f(sinx)+sinx*f(cosx)   =cosx*√[(1-sinx)/(1+sinx)]+sinx*√[(1-cosx)/(1+cosx)]   =√[((cosx)^2*(1-sinx))/(1+sinx)]+   √[((sinx)^2*(1-cosx))/(1+cosx)]   =√[((1-(sinx)^2)*(1-sinx))/(1+sinx)]+   √[((1-(cosx)^2)*(1-cosx))/(1+cosx)]   =√[((1+sinx)*(1-sinx)*(1-sinx))/(1+sinx)]+   √[((1+cosx)*(1-cosx)*(1-cosx))/(1+cosx)]   =√[(1-sinx)^2]+√[(1-cosx)^2]   =1-sinx+1-cosx   A>0,w>0,a∈【0,2π】   g(x)=2-sinx-cosx   =√2sin(x-3π/4)+2   (2)由-π/2+2kπ