若根号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)=2sinxcosx+2√3cos²x-√3=2sinxcosx+√3(2cos²x-1)=sin2x+√3cos2x=2sin(2x+π/3)最小正周期T=2π/2=π,

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

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

-根号3sinx^2+sinxcosx=-根号3*(1-cos2x/2)+1/2sin2x=根号3/2cos2x-根号3+1/2sin2x=sin(2x+60)-根号3/2f(a/2)=sin(a+6

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

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

主要用了4个公式cos^2x=(1+cos2x)/2sin^2x=(1-cos2x)/2sin2x=2sinxcosxasinx+bsinx=根号(a^2+b^2)sin(x+arctan(b/a))

(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

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

(sqrt是开方)f(x)=5sqrt(3)[cos(2x)]^2+sqrt(3)[sin(2x)]^2-4sin(x)cos(x),由于[cos(2x)]^2+[sin(2x)]^2=1,且2sin

f(x)=5√3*cos²x+√3*sin²x+4sinx*cosx-3√3=4√3*cos²x+4sinx*cosx-2√3=2√3*(2cos²x-1)+2

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

原式等于（3sinXcosX+cos²x-sin²x）/（sin²x+cos²x）再同时除以cos²x就行了

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