高三函数问题(过程)

∵f(sinx)=3-cos2x=3-(1-2sinx*sinx)
∴f(x)=3-(1-2x*x)
=2+2x*x

或者：设sinx=t
cos2x=1-2(sinx)^2=1-2t^2
f(sinx)=3-cos2x
f(t)=3-(1-2t^2)=2t^2+2
把t改成x
f(x)=2x^2+2
f(cosx)=2cos^2 x+2=1+cos2x+2=3+cos2x

选C

B

f(sinx)=2+1-cos(2x)=2+2sin^2 x
f(cosx)=2+2cos^2 x=3+2cos^2 x-1=3+cos2x

f(sinx)=3-cos2x=3-[1-2(sinx)^2]=2+2(sinx)^2
所以f(x)=2+2x^2
所以f(cosx)=2+2(cosx)^2=2+2(cosx)^2-1+1=3+cos2x
选C