已知函数f(x)=1-cos^2+2根号3sinxcosx-1/2cos2x,x属于R

原式=1-(1+cos2x)/2+√3sin2x-1/2cos2x. x∈R. =√3sin2x-cos2x+1/2. =2[(√3/2sin2x-1/2cos2x+1/2.∴f(x)=2sin(2x-π/6)+1/2.f(x)的最小证周期T=2π/2=π; 当sin(2x-π/6)=1时,f(x)max=2*1+1/2=5/2；当sin(2x-π/...

(2)若x。(0≦x。≦π/2）为f(x)的一个零点，求sin2x。的值。解：(1)。f(x)=(1-cos2x)/2+(√3)sin2x-(1/2)cos2x=(√3)sin2x-cos2x+1/2 =2[(√3/2)sin2x-(1/2)cos2x]+1/2=2[sin2xcos(π/6)-cos2xsin(π/6)]+1/2 =2sin(2x-π/6)+1/2；故最小正周期T...

f(x)=(1/2)(cosx)^2-3sinxcosx-(1/2)(sinx)^2 +1 ==(1/2)[(cosx)^2-(sinx)^2]-(1/2)2sinxcosx -sin2x =(1/2)(cos2x+sin2x)=(√2/2)[cos2xcos(pi/4)+sin2xsin(pi/4)]=(√2/2)cos(2x-pi/4)1)周期T=2pi/2=pi 2)对称轴2x-pi/4=kpi--->x=kpi/2+pi/...

f(x)=√3sinxcosx-1/2cos2x ＝（√3sin2x）/2-1/2*cos2x ＝sin2xcos30°-sin30°cos2x ＝sin(2x-30°）0<=x<=90°时，0<=2x<=180° -30<=2x-30<=150° 故最大值为1 最小值为 -√3/2

f（x）=1/(3/2*sin2x-cos2x-2)(3/2)^2+1^2再开方；既是根号13比上2；以下用a代替；f(x)=asin(2x-b)分之1（b为角）；最小周期就是1/π；最小值就是sin值为-1的时候，即为-1/a;

f(x)＝√3sinxcosx-1/2cos2x ＝√3/2sin2x-1/2cos2x ＝sin(2x-π/6).(1)最小值：f(x)|min＝-1,此时2x-π/6＝2kπ-π/2,即x＝kπ-π/6 (k为整数);最小正周期：T＝2π/2＝π.(2)f(C)＝1,则 sin(2C-π/6)＝1,即C＝π/3.R＝c/(2sinC)＝√3/(2·√3/2)...

【参考答案】f(x)=(1/2)×[(1+cos2x)/2]-(√3 /2)sin2x-(1/2)×[(1-cos2x)/2]+1 =(1/4)+(1/4)cos2x-(√3 /2)sin2x-(1/4)+(1/4)cos2x+1 =(1/2)cos2x-(√3 /2)sin2x+1 =sin(π/6- 2x)+1 =1-sin(2x -π/6)第一题：最小正周期是T=2π/2=π 当...

1/2cos^2x=1/2*(1+cos2x)*1/2=1/4+1/4cos2x √3/2sinxcosx=√3/4six2x 所以f(x)=√3/4six2x+1/4cos2x+5/4 再利用辅助角公式,f(x)=1/2*(√3/2six2x+1/2cos2x)+5/4 =1/2*sin(2x+π/6)+5/4 所以ω=2 t=2π/ω=π 所以最小正周期为π.

当x+π/3=π/2+2kπ;即：x=π/6+kπ时，f(x)取最大值，f(x)(MAX)=1 当x+π/3=-π/2+2kπ;即：x= - 5π/6+kπ时，f(x)取最小值，f(x)(min)= - 1 单调增区间的方法 是把(x+π/3)解代入到标准正弦函数中去解出，x 即为求；由-π/2+2kπ≤x+π/3≤π/2+...

f(x)= √3sinxcosx-(cosx)^2+1/2 = √3/2sin2x -1/2cos2x = sin(2x-π/6)则递减区间为 π/2 + 2kπ≤2x-π/6≤3π/2 + 2kπ 即f(x)的递减区间为 π/3+kπ≤ x ≤5π/6 +kπ