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

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

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

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

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

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