已知函数f(x)=asin(πx+α)+bcos(πx+β),其中a,b,α,β都是非零实数.f(2005)=-1,则f(2006)等于多少

f(x)=asin(πx+α)+bcos(πx+β)
f(2005)=asin(2005π+α)+bcos(2005π+β)=asin(π+α)+bcos(π+β)=-1
asin(π+α)+bcos(π+β)=-1
asin(α)-bcos(β)=-1
bcos(β)=asin(α)+1
f(2006)=asin(2006π+α)+bcos(2006π+β)=asin(α)+bcos(β)=asin(α)+asin(α)+1
=2asin(α)+1