Question

已知f(x)=asin(πx/2+α)+bcos(πx/2+β)+4,若f(2009)=8,则f(2011)=

Answer 1

f(2009)=asin(2009π/2+α)+bcos(2009π/2+β)+4=8

asin(2009π/2+α)+bcos(2009π/2+β)=4

f(2011)=asin(π+2009π/2+α)+bcos(π+2009π/2+β)+4=-asin(2009π/2+α)-bcos(2009π/2+β)+4=-4+4=0