√3sinxcosx+sin²x-1 2

f(x)=cos²x+2√3sinxcosx-sin²x=cos2x+√3sin2x=2[1/2cos2x+√3/2sin2x]=2(sinπ/6cos2x+cosπ/6sin2x

f(x)=-√3sin²x+sinxcosx=(1/2)*sin2x+(√3/2)*cos2x-√3/2=sin(2x+π/3)-√3/2所以函数f(x)的最小正周期是T=2π/2=π最大值

f(x)=3/2-3sin^2x-√3sinxcosx=3/2-3*(1-cos2x)/2-(√3/2)*sin2x=(3/2)*cos2x-(√3/2)*sin2x=√3*[(√3/2)*cos2x

f(x)=-√3sin^2x+sinxcosx=-√3(1-cos2x)/2+sin2x/2=sin2x/2+√3cos2x/2-√3/2=sin(2x+π/3)-√3/2因为X∈[0,π/2],所以

先化简2√3sinxcosx＝√3sin2x2sin^2x=-(1-2sin^2x)+1=-cos2x+1所以y＝√3sin2x-cos2x－1＝2（sin2xcos30-cos2xsin30）-1=

y=(sin²x+cos²x)+(√3/2)(2sinxcosx)+cos²x=1+(√3/2)sin2x+(1+cos2x)/2=(√3/2)sin2x+(1/2)co

f(x)=2sin^2x-2cos^2x+4√3sinxcosx-3=-2cos2x+2√3sin2x-3=4(-1/2*cos2x+√3/2sin2x)-3=4sin(2x-π/6)-3(1)求f(

f(x)=-√3sin2x+1/2sin2x+√3/2=(-√3+1/2)sin2x+√3/2因为sin2x的最小正周期为2π/2=π因为-√3+1/2小于0所以单调递增区间为（π/4+kπ,3π/4

sin(2x+π/3)-√3sin²x+sinxcosx+√3/2=sin(2x+π/3)-2sinx【（√3/2）sinx-（1/2）cosx】+（√3/2）=sin(2x+π/3)-2s

y=3sin²x-2√3sinxcosx+3cos²x+2cos²x=1-√3sin2x+cos2x-1+1=cos2x-√3sin2x+2=2(cosπ/3*cos2x

f(x)=3sin^2x+2根号3sinxcosx+5cos^2x=3sin^2x+√3sin2x+5cos^2x-4cos^x-4sin^2x+4=2sin(30+2x)+4T=∏最大值为62sin

f(x)=(√3sinx+cosx)²=4(√3/2*sinx+cosx*1/2)²=4(sinx*cos(π/6)+cosx*sin(π/6))²=4sin²

f(x)=-√3sin²x+sinxcosx=-√3(1-cos2x)/2+sin2x/2=-√3/2+√3/2cos2x+sin2x/2=-√3/2+sin(2x+π/3)∵x∈[0,π/

f(x)=-√3sin²x+sinxcosx=-√3(1-cos2x)/2+1/2*sin2x=√3/2*cos2x+1/2*sin2x-√3/2=cos2xcosπ/6+sin2xsinπ

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

1、f(x)=3sin��x+2√3sinxcosx+5cos��x=3(1-cos2x)/2+√3sin2x+5(1+cos2x)/2=3/2-3/2cos2x+√3sin2x+5/2+5/2cos

(a²＋b²＋c²)/(a²＋b²－c²)＝c/(2a＋c)利用余弦定理可得：2accosB/2abcosC＝c/(2a－c)即：cosB/

你可能是忙中出错了!题目应该是这样的：化简：f（x）＝2cosxcos（π/6－x）－√3（sinx）^2＋sinxcosx.［解］f（x）＝2cosx［cos（π/6）cosx＋sin（π/6）si

f(x)=(cosx)^4-(sinX)^4+2√3sinxcosx=[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]+√3sin2x=(cosx)^2-(sinx)^2+