Question

设f(x)=asin(πx+Q)+bcos(πx+B)+4,且f(2003)=5,则f(2004)＝

Answer 1

f(2003)=asin(2003π+Q)+bcos(2003π+Q)+4=5

则asin(2003π+Q)+bcos(2003π+Q)=1

f(2004)=asin(2003π+Q+π)+bcos(2003π+Q+π)+4

=-asin(2003π+Q)-bcos(2003π+Q)+4

=4-[asin(2003π+Q)+bcos(2003π+Q)]

=4-1

=3