F(X)=SIN(πx/2+α),且f(2010=1)则f(2012=?)

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F(X)=SIN(πx/2+α),且f(2010=1)则f(2012=?)F(X)=SIN(πx/2+α),且f(2010=1)则f(2012=?)F(X)=SIN(πx/2+α),且f(2010=1

F(X)=SIN(πx/2+α),且f(2010=1)则f(2012=?)
F(X)=SIN(πx/2+α),且f(2010=1)则f(2012=?)

F(X)=SIN(πx/2+α),且f(2010=1)则f(2012=?)
利用周期性可以了.
ƒ(x) = sin(πx/2 + α)
ƒ(2010) = sin(1005π + α)
= sin(251 * 4π + π + α) = sin(π + α) = - sin(α) = 1
==> sin(α) = - 1
ƒ(2012) = sin(1006π + α)
= sin(251 * 4π + 2π + α)
= sin(2π + α)
= sin(α)
= - 1

因为f(x)=asin(πx+α)bcos(πx+β),f(2010)=-1,
所以-1=asin(2010π+α)bcos(2010π+β)=absinαcosβ,
f(2012)=asin(2012π+α)bcos(2012π+β)=absinαcosβ=-1.

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