设f(sin x)=cos2x+1,求f(cos x).

问题描述:

设f(sin x)=cos2x+1,求f(cos x).

原式=1-2(sinx)^2+1
=2-2(sinx)^2
所以f(cosx)=2-2(cosx)^2

f(cos x)=f(sin(90`-x))=cos(180`-2x)+1=1-cos2x

f(sin x)=cos2x+1
f[sin(π/2-x)]=cos[2(π/2-x)]+1
f(cosx)=cos(π-2x)+1=1-cos2x=2sin^2 x

sin(pi/2-x)=cosx
f(sin x)=cos2x+1
f(cos x)=f(sin(pi/2-x))=cos2(pi/2-x)+1=1-cos2x

f(sinx)=cos2x+1
=1-2sin^2 x+1
=2-2sin^2 x
f(cosx)=2-2cos^2 x
=2(1-cos^2 x)
=2sin^2 x