Question

高一数学填空

1.y=5sin(2x+π/12)+4cos(2x+5π/12)的最大值。
2.x属于[π/6,2π/3]则函数y=2sin(2x-π/6)的至于为........递增区间是........

Answer 1

1.
y=5sin(2x+π/12)+4cos(2x+5π/12)
=5sin(2x+π/12)+4sin(2x-π/12)
=5sin2xcos(π/12)+5cos2xsin(π/12)+4sin2xcos(π/12)-4cos2xsin(π/12)
=9cos(π/12)*sin2x+sin(π/12)*cos2x

最大值={[9cos(π/12)]^2+[sin(π/12)]^2}^(1/2)
={80[cos(π/12)]^2+1}^(1/2)
=[40*(1+cos(π/6))+1]^(1/2)
=根号(41+20(根号3)]

2.
π/6<=x<=2π/3
π/6<=2x-π/6<=7π/6

值域：[-1,2]
递增区间是:[π/6,π/3]