f(2012)=asin(2012π+α)+bcos(2012π+β)+4
=asinα+bcosβ+4
=6
所以asinα+bcosβ=2
f(2013)
=asin(2013π+α)+bcos(2013π+β)+4
=asin(π+α)+bcos(π+β)+4
=-asin(α)-bcos(β)+4
=-2+4
=2
f(2012)=asin(2012π+α)+bcos(2012π+β)+4
=asinα+bcosβ+4
=6
所以asinα+bcosβ=2
f(2013)
=asin(2013π+α)+bcos(2013π+β)+4
=asin(π+α)+bcos(π+β)+4
=-asin(α)-bcos(β)+4
=-2+4
=2