关于同角三角函数的问题f(cosx)=x/2,求f(cos(4π/3))

问题描述:

关于同角三角函数的问题f(cosx)=x/2,求f(cos(4π/3))
1.f(cosx)=x/2 (0

1、cos(4π/3)=cos(2π-2π/3)=cos(2π/3)所以f(cos(4π/3))=f(cos(2π/3))=(2π/3)/2=π/32、f(sinx)=3-cos2x=3-(1-2(sinx)^2)=2+2(sinx)^2f(cosx)=f(根号1-(sinx)^2) ...