要使f(x)是与x无关的定值则要求cos^2(x+a)+cos^2 (x+b) =sin^2 x +k (k为常数)cos^2 x cos^2 a +sin^2 x sin^2 a +2cosxsinx cosasina +cos^2 x cos^2 b +sin^2 x sin^2 b+2cosx sinx cosbsinb =sin^2 x +kcos^2 x(c
f(x)=psinwx*coswx-coswx=(p/2)sin2wx-(1/2)cos2wx -1/2=(1/2)√(p+1)sin(2wx-φ) -1/2,其中 tanφ=1/p,φ为锐角1.由最大值为1/2,得(1/2)√(p+1)-1/2=1/2,p+1=4,p=√3,从而φ=π/6.由最小正
(1)f(x)=cosωx+2√3cosωxsinωx-sinωx =cos2ωx+√3sin2ωx =2sin(2ωx+π/6)y=sinωx相邻两条对称轴的距离等于 1/2 周期f(x)相邻两对称轴间距离不小于拍π/2,即周期不小于π.∴2ω≤2π/π→ ω≤1 展开 作业帮用户 2017-09-01 举报
sin2x=2sinx cosx 即sin2x=sinx cosxcos2x =cosx -sinx不懂追问数理化全精通为您解答望采纳O(∩_∩)O谢谢
已知函数f(x)=coswx+2根号3coswx*sinwx-sinwx(w>0),且f(x)相邻两对称轴间的距离不小于π/2(1)求w的取值范围(2)在△ABC中,角A、B、C所对边分别为a、b、c,且a=根号3,b+c=3,若当w取最大值时,f(A)=1,求
f(x)=cos(x+π/6)-sin(x+π/6)倍角公式 =cos[2(x+π/6)]=cos(2x+π/3)
f(x)=cosx+2tsinxcosx-sinx=(cosx-sinx)+t(2sinxcosx)=cos2x+tsin2x,根据题意,f'(x)=-2sin2x+2tcos2x≥0在区间(π/12,π/6]上恒成立,即t≥tan2x在区间(π/12,π/6]上恒成立,由于在区间(π/12,π/6]上,√3/3
f(x)=√3(cosx-sinx)-2sinxcosx= √3cos2x-sin2x= 2{cos2xcosπ/6-sin2xsinπ/6)= 2cos(2x+π/6)最小正周期 2π/2 = πx∈[0,2π]时2x∈[0,4π]2x+π/6∈[π/6,4π+π/6]f(x)=2cos(2x+π/
f(x)=cosx-sinx+2sinx=1-sinx -sinx+2sinx= -2sinx+2sinx+1=-2(sinx-1/2)+3/2最大值为当sinx=1/2时,f(x)=3/2最小值为当sinx=-1时,f(x)=-3
把sin(x+pi/6)拆开