f[y(x)]=1+cosx.y(x)=sinx/2,求f(x).
问题描述:
f[y(x)]=1+cosx.y(x)=sinx/2,求f(x).
答
cosx=cos(2*x/2)=1-2[sin(x/2)]^2
令t=sin(x/2)
f(t)=1+1-2t^2=2-2t^2
f(x)=2-2x^2