函数sinx cosx在区间[-2分之π

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

应该是f(x)=sin^2x+√3sinxcosx=-(-2sin^2x\2)+(√3\2)sin2x=-[(1-2sin^2x)\2]+(√3\2)sin2x+1\2=-(1\2)cos2x+(√3

公式cos2x=1-2sin²x,可以知道sin²x=(1-cos2x)/2后面的√3sinxcosx=√3sin2x/2所以原式=-(cos2x)/2+(√3sin2x)/2+1

f(x)=根号3sin2x+cos2x=2（cosπ/6sin2x+sinπ/6cos2x）=2sin（2x+π/6）因为函数在区间[0,π/2]上所以π/6≤2x+π/6≤7π/6当2x+π/6=π

f(x)=sin∧2x+√3sinxcosx=-1/2cos2x+√3/2sin2x-1/2再答：=sin（2x-π/6）-1/2再答：x∈［π/4,π/2］所以2x-π/6∈［π/3,5π/6］所以

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

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

f(x)=(1-cos2x)/2+√3/2*sin2x=√3/2sin2x-1/2*cos2x+1/2=sin2xcosπ/6-cos2xsinπ/6+1/2=sin(2x-π/6)+1/2π/4

f(x)=1+sin2x+2cos^2x=1+sin2x+1+cos2x=√2sin(2x+π/4)+22kπ-π/2≤2x+π/4≤2kπ+π/2kπ-3π/8≤x≤kπ+π/8[0,π/8]

f(x)=√3sinxcosx+cos²x+a=(sin2x)×(√3/2)+(cos2x+1)×(1/2)+a=(cosπ/6)×sin2x+(sinπ/6)×cos2x+1/2+a=si

f(x)=根号3/2*sin2x-1/2cos2x=cospi/6sin2x-sinpi/6cos2x=sin(2x-pi/6)f(0)=-1/2f(pi/4)=根号3/2函数值的范围[-1/2,根号

f(x)=sin²x+√3sinxcosx=(1-cos2x)/2+(√3/2)sin2x=(√3/2)sin2x-(1/2)cos2x+1/2=sin(2x-π/6)+1/2x∈[π/4,

（2）1-3

y=sin²x+√3sinxcosx=(1-cos2x)/2+√3/2sin2x=√3/2sin2x-1/2cos2x+1/2=cosπ/6sin2x-sinπ/6cos2x+1/2=sin

F[x]=sinxcosx+cos^2x-1/2=1/2sin2x+1/2(cos2x+1)-1/2=1/2(sin2x+cos2x)=√2/2sin(2x+π/4)最小正周期T=2π/W=π2x+π

y=[(sinx)^4-(cosx)^4]+[(2根号3)*(sin2x)/2]=[(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2]+(根号3)sin2x=[(根号3)sin

最大值是：sinA+(√3)/2f(x)=sinA+(√3)sinxcosxf(x)=sinA+[(√3)/2]sin(2x)f'(x)=(√3)cos(2x)因为：x∈[π/4,π/2]所以：2x∈

f(x)=(1-cos2x)/2+根号3/2sin2x=1/2+sin(2x-π/6)x∈【π/4,π/2】,2x-π/6∈【π/3,5π/6】,sin(2x-π/6)∈【1/2,1]f(x)max=

f(x)=sinxcosx-m(sinx+cosx)=1/2sin2x-m(sinx+cosx)求导得到f’（x）=cos2x+m（sinx-cosx）函数f(x)在区间(0,π/2)是单调减函数所以

f(x)=2√3sinxcosx+2cos²x+m=√3sin2x+1+cos2x+m=2sin(2x+π/6)+m+1.0再问：在三角形ABC中角ABC所对的边长abc若F（A）=1，si