若根号下3sinx cosx=4-m

(1)f(x)=sinxcosx+√3cos²X-√3/2=sin2x/2+√3cos2x/2+√3/2-√3/2=sin(2x+π/3).(2)f(x)的最小正周期为π,值域是[-1,1]

f(x)=1/2sin2x-√3/2(1-cos2x)=1/2sin2x+√3/2cos2x-√3/2=sin(2x+π/3)-√3/21、最小正周期T=2π/2=π2、0≤x≤π/2π/3≤2x+π

f(x)=5cos²x+sin²x-4√3sinxcosx=4cos²x+cos²x+sin²x-2√3sin2x=2(cos2x+1)+1-2√3s

y=√3*sinxcosx+cos^2x=√3/2sin2x+(1+cos2x)/1=√3/2sin2x+1/2*cos2x+1/2=sin(2x+π/6)+1/2周期T=2π/2=π2Kπ-π/2≤

f(x)=(√3)sinxcosx+cos2x+1f(x)=(√3)(2sinxcosx)/2+cos2x+1f(x)=(√3/2)sin2x+cos2x+1f(x)=(√7/2)[(√3/2)(2/

f(x)=2√3sinxcosx-cos2x=√3sin2x-cos2x=2(sin2x*√3/2-cos2x*1/2)=2sin(2x-π/6)x=π/12;函数f(x)的图象可以由函数y(x)=2

f(x)=2cosx[1/2sinx+√3/2cosx]-√3sin^2(x)+sinxcosx=sinxcosx+√3cos^2(x)-√3sin^2(x)+sinxcosx=sin2x+√3cos

f(x)=cos^4x-2根号3sinxcosx-sin^4x=f(x)=(cos^2x-sin^2x)(cos^2x+sin^2x)-2根号3sinxcosx=cos2x-根号3sin2x=-2si

f(x)=5根号下3cos^2x+根号下3sin^2x-4sinxcosx=4根号下3cos^2x-4sinxcosx+根号下3=2根号下3(2cos^2x-1)-2sin2x+3根号下3=2根号3c

y=1-4(sinx)^2-2倍根号3sin（2x）=1-2（1-cos2x）-2倍根号3sin（2x）=4sin（π/6-2x）-1当x∈[0,π/2],π/6-2x属于[-5π/6,π/6]所以最

f(x)=5根号下3cos^2x+根号下3sin^2x-4sinxcosxx属于[π/4,7π/24]f(x)=根号3(5(1+cos2x)/2)+根号3(1-cos2x)/2-2sin2xf(x)=

(sinX+cosX)平方=2所以sinX平方+cosX平方+2sinXcosX=2因为sinX平方+cosX平方=1所以sinXcosX=0.5

怎么感觉cosx应该是平方啊再问：嗯的，打错了再答：(cosx)^2=(1+cos2x)/2sinxcosx=1/2*sin2x所以原式=(1+cos2x)/2-根号3/2*sin2x+1=1/2*c

f(x)=cos2x+(√3)sin2x=2cos(2x-π/3)故Tmin=2π/2=π单增区间：由-π+2kπ≤2x-π/3≤2kπ,-2π/3+2kπ≤2x≤2kπ+π/3,得-π/3+kπ≤x

答：f(x)=√3sinxcosx+(cosx)^2=(√3/2)*2sinxcosx+(1/2)*[2(cosx)^2-1]+1/2=(√3/2)sin2x+(1/2)cos2x+1/2=sin(2

f(x)=sin^4x-2根号3sinxcosx-cos^4x+1=(sin^2x-cos^2x)(cos^2x+sin^2x)-2根号3sinxcosx+1=-cos2x-根号3sin2x+1=-2

f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1

cos2x-2根号3sinxcosx=cos2x-根号3（2sinxcosx）运用倍角公式得=cos2x-根号3sin2x运用辅助角公式得=-2sin(2x-六分之π)由ω=2,T=二派除以ω,所以周

解:原式=√3sin2x＋cos2x＋1=2(√3/2sin2x＋1/2cos2x＋1=2cos(2x-pai/3)＋1.

不对,因为f（x）=cos2x-2√3sinxcosx=cos2x-√3sin2x=2[sinπ/6cos2x-cosπ/6sin2x]=2sin(π/6-2x)左移5π/12,f（x）=2sin【π