设f(1)=a,f(x)=asin(πx+α)+bcos(πx+α)其中abα∈R且a b ≠0,α≠kπ（k∈z）若f(2009)=5,求f(2010)

f(x)=asin(πx+α)+bcos(πx+α)
f(1)=asin(π+α)+bcos(π+α)=-（asinα+bcosα）=a
f(2009)=asin(2009π+α)+bcos(2009π+α))=asin(π+α)+bcos(π+α)
=-（asinα+bcosα）=a=5
f(2010)=asin(2010π+α)+bcos(2010π+α)=asinα+bcosα=-a=-5