1.f（2009）=a*sin(2009*π+A）+b*cos(2009π+B) =a*sin(π+A）+b*cos(π+B)
=-a*sinA-b*cosB=-1,則a*sinA+b*cosB=1
f(2010)= a*sin(2010*π+A）+b*cos(2010π+B) =a*sin(2π+A）+b*cos(2π+B)
=a*sinA+b*cosB
=1
2.y=2sin(π/6-2x)=-2sin(2x-π/6)
在一個周期內(nèi),求導并等于0,y’=-4cos(2x-π/6)=0,
2x-π/6=π/2,2x-π/6=3π/2,
解得：x=π/3,x=5π/6
求二階導數(shù),y’’=8sin(2x-π/6)
當x=π/3時,y’’0,有極小值；
所以,函數(shù)的單調(diào)遞增區(qū)間是：x∈[0,π/3]U[5π/6,π]