已知函数f x 2sin wx π 6 a(w>0)与g x 2cos 2x

当x∈[0,π/2]时,-1/2

(1)sin(wx+π/6)=sinwxcosπ/6+coswxsinπ/6sin(wx-π/6)=sinwxcosπ/6-coswxsinπ/6f(x)=sin(wx+π/6)+sin(wx-π/6

1.f(x)=sin(2x+π/6)+sin(2x-π/6)+cos2x+a=√3/2sin2x+1/2cos2x+√3/2sin2x-1/2cos2x+cos2x+a=√3sin2x+cos2x+a

第一个问题：f（x）＝sin2xcos（π/6）＋cos2xsin（π/6）＋sin2xcos（π/6）－cos2xsin（π/6）－cos2x＋a＝2sin2xcos（π/6）－cos2x＋a＝2［

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+af(x)=根3sinx+cosx+af(x)=2sin(x+π/6)+a因为f(X)最大值为1所以2+a=1a=-1

再问：谢啦w

1、f(x)=sinxcosπ/6+cosxsinπ/6+sinxcosπ/6-cosxsinπ/6+cosx+a=2sinxcosπ/6+cosx+a=√3sinx+cosx+a=√[(√3)&su

f(x+pai/6)=Asin(2x+pai/3+a)=Acos(pai/6-a-2x)pai/6-a=2kpai,pai/6-a=2kpai+paif(x)=Asin(2x+pai/6-2kpai)

1、f(x)=sin(2x-π/6)-2cos²x=(√3/2)sin2x-(1/2)cos2x-(1+cos2x)=(√3/2)sin2x-(3/2)cos2x-1=√3sin(2x-π/

y=cosB+6sinay=cos（π-2a)+6sina=-cos2a+6sina=-(1-2sin²a)+6sina=2sin²a+6sina-1=2(sina+3/2)&su

∵sin(3x+π6)的最大值为1，最小值为-1．∴当b＞0时，函数y＝a−bsin(3x+π6)的最大值为a+b=5，最小值为a-b=1．解之得a=3，b=2．当b＜0时，函数y＝a−bsin(3x

fx=(x-a)lnxf'(x)=lnx+(x-a)/x函数在（0,+无穷）上为增函数∴f'(x)=lnx+(x-a)/x>=0lnx+1-a/x>=0lnx+1>=a/x∵x>0∴xlnx+x>=a

f(x)=sin(x+π/6)+sin(x-π/6)+cosx+a=sinx*cos(π/6)+cosx*sin(π/6)+sinx*cos(π/6)-cosx*sin(π/6)+cosx+a=（√3

首先得T/2=2π-3π/4=5π/4所以：T=5π/2,即2π/w=5π/2,所以：w=4/5；所以：y=sin(4x/5+A),把点(3π/4,-1)代入,得：-1=sin(-3π/5+A)所以：

最小正周期=π；当：2kπ-π/2≤2x-π/6≤2kπ+π/2kπ-π/6≤x≤kπ+π/3所以增区间为：[kπ-π/6,kπ+π/3]3.x∈{0,π/2}时,-π/6≤2x-π/6≤5π/6f(

-1

（1）f（x）=a*b=2sin（x+π/6)+2cosx=√3sinx+3cosx化成同名函数f（x）=2√3sin（x+π/6)(提取系数平方和)则2kπ-π/2≤x+π/6≤2kπ+π/2解得f

比起跟你讲答案,我更愿意跟你讲思路：对于第一题,可以通过替换将2x+pi/6看成一个变量a,函数的单调区间显然只由sinx函数决定,所以可以令-pi/2+2kpi

-pi/2-2pi/3pi/32.0->pi/33.a=04.f(pi/2)=-2sinpi/6+a=-2a=sqrt(3)-2

（1）f（x）=4cosx•sin（x+π6）+a=23sinxcosx+2cos2x+a=3sin2x+cos2x+1+a=2sin（2x+π6）+1+a，∵sin（2x+π6）≤1，∴f（x）≤2