已知f(sin x)=cos 3x,求f(cos π/9)的值.

f(sin(pai/2-x))=cos[3(pai/2-x)]f(cosx)=cos(3pai/2-3x)f(cospai/9)=cos(3pai/2-pai/3)=-sinpai/3=-根号3/2 f(cospai/9)=f(sin(pai/2+pai/9))=cos(3(pai/2+pai/9))=cos(3pai/2+pai/3)=cospai/3=根号3/2如果要分正负号,则要讨论:f(sinx)=cos3xf(sinx)=cos(2x+x)=cosxcos2x-sinxsin2x=cosx(1-2sin^2x)-2sinxcosxsinxf(sinx)=-2cosxsin^2x+cosx-2sin^2xcosx=cosx-4cosxsin^2x=cosx(1-4sin^2x)f(sinx)=根号(1-sin^2x)*(1-4sin^2x) cosx>=0。1f(sinx)=根号(1-sin^2x)(4sin^2x-1) cosx