若f(sinx)=sin3x,则f(cosx)=?A.-cos3xB.cos3xC.sin3xD.-sin3x

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若f(sinx)=sin3x,则f(cosx)=?A.-cos3xB.cos3xC.sin3xD.-sin3x
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若f(sinx)=sin3x,则f(cosx)=?A.-cos3xB.cos3xC.sin3xD.-sin3x
若f(sinx)=sin3x,则f(cosx)=?
A.-cos3x
B.cos3x
C.sin3x
D.-sin3x

若f(sinx)=sin3x,则f(cosx)=?A.-cos3xB.cos3xC.sin3xD.-sin3x
f(cosx)=f[sin(π/2-x)]
=sin[3(π/2-x)]
=sin(3π/2-3x)
=sin[π+(π/2-3x)]
=-sin(π/2-3x)
=-cos3x

怎么又来这道题

令sinx=u,则(cosx)^2=(1-u^2),
f(u)=sin3x=sin(2x+x)=sin2xcosx+cos2xsinx=2sinx(cosx)^2+(1-2(sinx)^2)sinx=2u(1-u^2)+(1-2u^2)u=3u-4u^3,
则f(cosx)=3cosx-4(cosx)^3.

3cosx-4*(cosx)^3

A
f(cosx)
=f[sin(π/2-x)]
=sin[3(π/2-x)]
=sin(3π/2-3x)
=-cos3x