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)
答
(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