f(sinx)=3-cos2x

如果f(sinx)=3-cos2x,则f(cos2x)等于多少?题目是cos2x,不是cosxf(sinx)=3-(1-2sin²x)=2+2sin²x所以f(x)=2+2x²f(cos2x)=2+2cos&s

设f(sinx)=3-cos2x,则f(cosx)=?令t=sinx则f(sinx)=f(t)=3-1+2sin²x=2+2t²,(-1≤t≤1)∴f(t)=2+2t²,(-1≤t≤1)∴f(cosx)=2+2

f(sinx)=3-cos2x则f(cosx)=?f(cosx)=f[sin(π/2-x]=3-cos[2(π/2-x)]=3-cos(π-2x)=3-(-cos2x)=3+cos2x

若f(sinx)=3-cos2x,则f(cosx)=?若f(sinx)=3-cos2x=2+1-cos2x=2+(sinx)^2+(cosx)^2-[(cosx)^2-(sinx)^2]=2+2(sinx)^2f(t)=2+2t^2t=si

若f(sinx)=3-cos2x,则f(x)=令sinx=Tcos2x=1-2sinx^2=1-2T^2f(sinx)=3-cos2xf(T)=3-{1-2T^2}所以f（x）=2x^2+2f(sinx)=3-cos2x=3-(1-2sin

若f(sinx)=3-cos2x,则f(cos)=?答：f(sinx)=3-cos2x=1-cos2x+2=2(sinx)^2+2所以：f(x)=2x^2+2所以：f(cosx)=2(cosx)^2+2=cos2x+1+2=3+cos2xc

f(sinx)=3-cos2x,则f(cosx)=?（详解本题的思路的是通过三角函数的转化达到解题目的利用cosx=sin(π/2-x)来解f(cosx)=f(sin(π/2-x))=3-cos[2*(π/2-x)]=3-cos(π-2x)

f(sinx)=3-cos2x,则f（cosx）=f(sinx)=3-cos2x=3-[1-2(sinx)^2]=2+2(sinx)^2,即f(sinx)=2+2(sinx)^2把其中的sinx换成cosx,得f(cosx)=2+2(cos

f（sinx)=3-cos2x则f(cosx)等于什么cos2x=1-2sinx^2=2cosx^2-1则f（sinx）=2+2sinx^2令t=sinx（-1

已知f(sinx)=3-cos2x求f(cosx)设sinx=t,则cos2x=1-2(sinx)^2=1-2t^2,所以f(t)=3-(1-2t^2)=2+2t^2所以f(cosx)=2+2(cosx)^2再化简一下就是：3+cos2x明

f（sinx）=3-cos2x,求f（cosx）倍角公式：cos2x=1-2sin²x所以：f(sinx)=3-(1-2sin²x)=2sin²x+2令sinx=t,则：f(t)=2t²+2所以,f(

设f(sinx)=cos2x,则f(1/3)等于?f(sinx)=cos2x=1-2(sinx)^2所以f(x)=1-2x^2f(1/3)=1-2*(1/3)^2=7/9f（sinx）=cos2xf（sinx）=1-2sinx^2f(x)=

f（sinx)=3-cos2x则f(cosx)等于什么f（sinx)=3-cos2x=3-(1-2sin²x)=2+2sin²x则：f(x)=2+2x²所以,f(cosx)=2+2cos²x=2+1+

已知f(sinx)=3-cos2x求f(cosx)解f(cosx)=f(sin(π/2-x))=3-cos2(π/2-x)=3-cos(π-2x)=3-[-cos2x]=3+cos2x

若f(cosX)=3-cos2X,则f(sinX)因为sinx^2=1-cos^2所以f(cosX)=3-cos2X=3-（2cos^2x-1）=2-2cos^2x令t=cosxf(t)=2-2t^2所以f(sinx)=2-2sin^2x=

f(sinx)=3-cos2x,则f(cos)的表达式f(cosx)=f[sin(π/2-x)]=3-cos[2(π/2-x)]=3-cos(π-2x)=3+cos2x∵cosx=sin（90°-x）∴f（cosx）=3-cos2（90°-

f（sinx)=3-cos2x则f(cosx)等于什么F（SINX)=3-cos2X,则F（COSX）等于对每一步进行解释说明,...原式=f[sin(π/2-x）】=3-cos2[(π/2-x)]=3-cos(π-2x)=3+cos2xf

f(sinx)=cos2xf'(x)=?f(sinx)=cos2x=1-2(sinx)^2letu=sinxf(u)=1-2u^2u=xsothatf'(x)=-4x-4xf(sinx)=cos2x=1-2(sinx)^2令t=sinxf(

F(sinx)=cos2x,求F(x)F(sinx)=cos2x=1-2sinx²所以F(x)=1-2x²x定义域是[-1,1]

如果f(sinx)=3-2cos2x,则f(cos2x)等于多少?f(sinx)=3-2cos2x=1+4(sinx)^2所以f(x)=1+4x^2从而有f(cos2x)=1+4(cos2x)^2