若f(cosX)=3-cos2X,则f(sinX)

问题描述:

若f(cosX)=3-cos2X,则f(sinX)

f(cosX)=3-cos2X
=3-(2cos^2x-1)
=2-2cos^2x
令t=cosx
f(t)=2-2t^2
所以
f(sinx)=2-2sin^2x
=1+(1-2sin^2x)
=1+cos2x
答题不易 望采纳

令cosX=t,则f(cosX)=f(t)=3-(2t^2-1),令t=sinx,则f(sinx)=3-[2(sinx)^2-1]
所以f(sinx)=3+[1-2(sinx)^2]=3+cos2x

因为sinx^2=1-cos^2
所以f(cosX)=3-cos2X =3-(2cos^2x-1) =2-2cos^2x 令t=cosx f(t)=2-2t^2所以 f(sinx)=2-2sin^2x =1+(1-2sin^2x) =1+cos2x