设f(x)=asin(πx+Q)+bcos(πx+B)+4,且f(2003)=5,则f(2004)=
问题描述:
设f(x)=asin(πx+Q)+bcos(πx+B)+4,且f(2003)=5,则f(2004)=
答
f(2003)=asin(2003π+Q)+bcos(2003π+Q)+4=5则asin(2003π+Q)+bcos(2003π+Q)=1f(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