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+B)+4
=asin(π+Q)+bcos(π+B)+4
=-asinQ-bcosB+4=5
所以asinQ+bcosB=-1
f(2004)=asin(2004π+Q)+bcos(2004π+B)+4
=asinQ+bcosB+4
=-1+4
=3
答
f(2003)=asin(2003π+Q)+bcos(2003π+B)+4=5
asin(π+Q)+bcos(π+B)+4=5
-asinQ-bcosB+4=5
asinQ+bcosB=-1
f(2004)=asin(2004π+Q)+bcos(2004π+B)+4
=asin(Q)+bcos(B)+4
=-1+4
=3
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