设函数f(x)=αsin(πx+α)+bcos(πx+β),且f(2011)=2012,求f(2012)的值

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设函数f(x)=αsin(πx+α)+bcos(πx+β),且f(2011)=2012,求f(2012)的值
设函数f(x)=αsin(πx+α)+bcos(πx+β),且f(2011)=2012,求f(2012)的值

设函数f(x)=αsin(πx+α)+bcos(πx+β),且f(2011)=2012,求f(2012)的值
f(x)=αsin(πx+α)+bcos(πx+β)
所以f(x+1)=αsin(πx+π+α)+bcos(πx+π+β)=-αsin(πx+α)-bcos(πx+β)=-f(x)
所以f(2012)=-f(2011)=-2012

f(2011)=αsin(2011π+α)+bcos(2011π+β)=αsin(π+α)+bcos(π+β))=-αsin(α)-bcos(β)
f(2012)=αsin(2012π+α)+bcos(2012π+β)=αsin(α)+bcos(β))=-f(2011)=-2012

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