3=f(2011)=asin(2011π+α)+bcos(2011π+β)+4=-asinα-bcosβ+4
∴asinα+bcosβ=1
∴f(2012)=asin(2012π+α)+bcos(2012π+β)+4=asinα+bcosβ+4=1+4=5
已知函数f(x)=asin(πx+α)+bcos(πx+β)+4,x∈R,且f(2011)=3,则求f(2012)的值
已知函数f(x)=asin(πx+α)+bcos(πx+β)+4,x∈R,且f(2011)=3,则求f(2012)的值
数学人气：454 ℃时间：2019-11-21 19:13:53
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