sin(2x π 6)在 (-π 3,π 4)上的单调性怎么看

2sin^2[(π/4)+x]+根号3（sin^x-cos^x）-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x=√3-1/2sin2x-√3/2cos2x=√3-sin(2x+π/3)∴最小正周期T=2π/2=π单调增区间：π/2+2

y＝sin(x+π3)sin(x+π2)＝(sinxcosπ3+cosxsinπ3)cosx=12sinxcosx+32cos2x＝14sin2x+32•1+cos2x2=34+12sin(2x+π3

令a=π-x则a趋于0sin3x=sin(3π-3a)=sin3asin2x=sin(2π-2a)=-sin2a所以原式=-lim(a→0)sin3a/sin2asin3a和sin2a的等价无穷小是3

你的题目是这样的吗已知函数f(X)=2sin(ax-π/6)sin(ax+π/3)(其中a为正常数,x∈R)的最小正周期为π（1）求a的值（2）在△ABC中,若A＜B,且f(A)＝f(B)＝1/2,求

f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-^3sin^2x+sinx*cosx=sin2x+√3cos2x=2si

sin(x+π/6)=1/3sin(5π/6-x)=sin[π-(x+π/6)]=1/3sin^2(π/3-x)=sin^2[π/2-(x+π/6)]=cos^2(x+π/6)=1-sin^2(x+π

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x=cos2x+1/2*cosx-根号3/2*sina+根号3/2*sinx+1/2*cosx+3sin^2x=cos2x+cosx

fx=2cosx(0.5sinx+根号3/2cosx)-根号3sin*2x+sinxcosx=2sinxcosx+根号3（cos*2x-sin*2x）=sin2x+根号3cos2x=2sin（2x+派

sin(2x+π/3)

f(x)=sin(2x-π/6)=cos[π/2-(2x-π/6)]=cos(2π/3-2x)=cos(2x-2π/3),故f(x)为偶函数,且其关于x=π/3对称,周期为π,因此a的最小整数为π.

原式=sin（x+π/3）+√3cos（x+π/3）+2sin（x-π/3）=2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3)=2sin(x+π/3+π/6)+2

f(x)=sinxcos(π/6)+cosxsin(π/6)+cos(2π/3)cosx+sin(2π/3)sinx化简得,f(x)=√3sinx(显然易见)f(x)在[-π/2,π/2]上是增函数

f(x)=2cos(x-π/6)sin(x+π/6)-√3*(sin(x-π/6))^2+sin(x-π/6)cos(x-π/6）=sin2x+sin(π/3)-(√3/2）[1-cos(2x-π/3

-cosx

令x=π/2-t,dx=-dt当x=0,t=π/2,当x=π/2,t=0L=∫(0-->π/2)e^sinx/(e^sinx+e^cosx)dx=∫(π/2-->0)e^sin(π/2-t)/[e^s

ln(3+x)/(3-x)是奇函数,∫[-π/2,π/2]sin^2(x)*ln(3+x)/(3-x)dx=0∫[-π/2,π/2](sinx)^6dx=2∫[0,π/2](sinx)^6dx有公式=

代入法求极限结果=0再问：不是0吧再答：诶看错了应该是1/2sin(pi/2-pi/3)-cos(pi/2)=1/2-0=1/2再问：真这么简单？？再答：这明显是代入法求极限啊。不然你准备怎么做？再问

y=sinx单调增区间[2kπ-π/2,2kπ+π/2](k∈z)所以sin(2x-π/6)的单调递增区间为2kπ-π/2≤2x-π/6≤2kπ+π/2,k∈Z-π/6+kπ≤x≤π/3+kπ,k∈Z