1.f(x)=√(a^2+b^2)sin(wx+arctan(b/a))周期为π,w=2π/π=2f(π/12)=4所以√(a^2+b^2)=4,a^2+b^2=16sin(2*π/12+arctan(b/a)=1,arctan(b/a)=π/2-π/6=π/3,b/a=√3解得a=2,b=2√3f(x)=2sin2x+2√3cos2x2.g(x)=2sin(π/3-2x)+2√3cos(π/3-2x)
=√3cos2x-sin2x+√3cos2x+3sin2x
=2√3cos2x+2sin2x
=4sin(2x+π/3)π/2+2kπ<=2x+π/3<=3π/2+2kπ，π/12+kπ<=x<=7π/12+kπ时，单调递减x∈[-5π/12+kπ,π/12+kπ]是单调递增 |
