y=√3sinωxcosωx-cos^2ωx 3 2最小正周期为Π,x=Π 6

f(x)=根号3cos²ωx+sinωxcosωx=根号3/2(cos²2ωx)+1/2(sin2ωx)=sin(2ωx+π/3)；y轴右侧的第一个最高点的横坐标为π/6,π/3ω

f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)=(1-cos2ωx)/2+(√3/2)sin2ωx=sin(2ωx-π/6)+1/2∵函数y=f(x)的图像相邻两条对称轴之间

cos2ωx=1-2sin²ωxf（x）=sin²ωx+√3cosωxcos（π/2-ωX）=(1-cos2ωx)/2+√3cosωxsinωx=1-(1/2)cos2ωx+(√3

1、(1)、y=√3/2sin2ωx-1/2cos2ωx+1=sin(2ωx-π/6)+1,T=2π/|2ω|=π,故|ω|=1,又当x=π/6时,函数有最小值,所以ω=-1.∴y=1-sin(2x+

(1)原式=根号3(1+cos2wx)/2+sin2wx/2+a=根号3cos2wx/2+sin2wx/2+根号3/2=sin(2wx+pi/3)+a+根号3/2求出单调递增区间为[kpi/w-5pi

1）这道题我刚做过,化简得f(x)=COS^（2wX-30"）所以w=0.52）a=根号3-1

f(x)=√3cosωx+sinωxcosωx+a=√3/2×(2cosωx-1)+√3/2+1/2×2sinωxcosωx+a=√3/2cos2ωx+1/2sin2ωx+a+√3/2=sin(π/3

f(x)=cos2wx+√3sin2wx+m+1=2sin(2wx+π/6)+m+1由题意得：2w*π/6+π/6=π/2∴w=1f(0)=m+2=2∴m=0∴f(x)=2cosx^2+2√3sinx

f(x)=√3cosωx+sinωxcosωx+a=√3/2×(2cosωx-1)+√3/2+1/2×2sinωxcosωx+a=√3/2cos2ωx+1/2sin2ωx+a+√3/2=sin(π/3

∵[xcos(x+y)+sin(x+y)]dx+xcos(x+y)dy=0==>xcos(x+y)dx+xcos(x+y)dy+sin(x+y)dx=0==>xcos(x+y)(dx+dy)+sin(

合并同类项么,很简单的只要你愿意去做左边=cos*x（cos*y+sin*y）+sin*x（cos*y+sin*y)=cos*x+sin*x=1=右边

原函数可以化为y=根号13sin(2wx+u)--根号31,绝对值X1--X2最小值为π/2,且w＞0,所以T/2=π/2=2π/2w,解得w=12,y=根号13sin(2a+u)--根号3=2/3嗯

f(x)=sin^2ωx-2√3sinωxcosωx-cos^2ωx+λ=-√3sin2wx-cos2wx+λ=2sin（2wx+7π/6）+λ又函数关于x=π对称故2wπ+7π/6=kπ+π/2（k

f(x)=√3/2cos²ωx+sinωxcosωx+a式子中cos²ωx是平方还是2倍呀,要是2倍就好算多了.1,先按2倍算一下,你看看.f(x)=√3/2cos²ωx

y=sin⁴3xcos³4xdy/dx=cos³4x*d(sin⁴3x)/dx+sin⁴3x*d(cos³4x)/dx=cos

y=sqrt(13)sin(2wx+A)-sqrt(3)2w(x1-x2)=2pi,A=arctan(2sqrt(3))w=2f(a)=sqrt(13)sin(4a+A)-sqrt(3)=2/34a+

f(x)=-√3sinωxcosωx+cos²ωx=-(√3/2)sin(2ωx)+[1+cos(2ωx)]/2=cos(2ωx)*cos(π/3)-sin(2ωx)*sin(π/3)+1/

（Ⅰ）f(x)＝√3cos2ωx＋sinωxcosωx＋a=√3cos2ωx＋(1/2)sin2ωx＋a=[(√3)^2＋(1/2)^2]sin[2ωx+arctan(√3/(1/2))]＋a=(13

（Ⅰ）函数f（x）=32-3sin2ωx-sinωxcosωx=32−3•1−cos2ωx2−12sin2ωx=32cos2ωx−12sin2ωx=−sin(2ωx−π3)．因为y=f（x）的图象的一

sin(x-y)=sinxcosy-cosxsiny,sin(x+y)=sinxcosy+cosxsinysin(x-y)sin(x+y)=sin²xcos²y-cos²