已知x属于(0,2π),sinx,cosx分别是方程x2-kx k 1=0

f(x)=2sin^2x+sin2x=2sin^2x-1+sin2x+1=-cos2x+sin2x+1=sin2x-cos2x+1=(√2)sin(2x-π/4)+1因为x∈[0,π]所以2x-π/4

sin(3π-x)-cos(5π+x)=1/2-sqr(3)/2sinx+conx=1/2-sqr(3)/2两边平方得1+2sinxconx=1-sqr(3)/2sin2x=-sqr(3)/2因为x属

(1)sin(wx+π/6)=sinwxcosπ/6+coswxsinπ/6sin(wx-π/6)=sinwxcosπ/6-coswxsinπ/6f(x)=sin(wx+π/6)+sin(wx-π/6

函数f(x)＝sinx＋sin(x＋π/2)=sinx＋cosx=√2[√2/2)sinx+(√2/2)cosx]=√2sin(x+π/4)x属于R,当sin(x+π/4)=1时最大,所以√2sin(

(sinx)^2表示sinx的平方(sinx)^2+2(sinx)^2cosx+(cosx)^2-(sinx)^2=12(sinx)^2cosx+(cosx)^2=(sinx)^2+(cosx)^22

y=4sin(2x+π/4)+1单调区间为：单调增：2kπ-π/2

tanx=1/2tan2x=2tanx/(1-tan²x)=4/3sinx/cosx=tanx=1/2cosx=2sinxsin²x+cos²x=1所以sinx=√5/5

（1）向量a=(cos3x/2,sin3x/2),向量b=(cosx/2),-sinx/2),向量a·b=(cos3x/2)*(cosx/2)+(sin3x/2)*(-sinx/2),=cos(3x/

2π/w=6π所以w=1/3x/3+φ=π/2+2kπ或x/3+φ=-π/2+2kπ（k属于z)φ=π/3+2kπ或φ=-5π/6+2kπ又-π

a=(cos3/2x,sin3/2x),b=(cosx/2,-sinx/2),a^2=1b^2=1,ab=cos3/2xcosx/2-sin3/2xsinx/2=cos(3/2x_x/2)=cos2x

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

[-0.5,0.5]

1/sinβ=(cosαcosβ-sinαsinβ)sinα整理得:(1+cosα*cosα)sinβ=2sinαcosαcosβ所以,tanβ=sinαcosα/(1+1-2(sinα)^2)/2=

x属于(0,π),因此sinx不等于0sin(π/2-2x)=cos2x=1-2(sinx)^2,cos(π+x)=-cosx,所以：√3sinx-(sin(π/2-2x))/(cos(π+x))*c

x∈[-2π/9,π/6]3x+π/3∈[-π/3,5π/6]sin(3x+π/3)∈[-√3/2,1]2sin(3x+π/3)∈[-√3,2]函数的最大值=2函数的最小值=-√3

F(x)=根号3(sin^2X-cos^2X)+2sinXcosX=√3cos2x+sin2x=2sin(2x+π/3)(1)X属于[0,2π/3]2x+π/3属于[π/3,5π/3]2sin(2x+

解(1)由f（x）=sin（π/3-2x）=-sin(2x-π/3)当2kπ-π/2≤2x-π/3≤2kπ+π/2,k属于Z时,y是减函数,即kπ-π/12≤x≤kπ+5π/12,k属于Z时,y是减函

tanx=3/4sin(π-x)=-3/5

提供一个思路吧,整理成二次方程的形式,就是利用二次方程的在定义域[0,1]最小值也大于0就行了,很好算的就是麻烦一点

比起跟你讲答案,我更愿意跟你讲思路：对于第一题,可以通过替换将2x+pi/6看成一个变量a,函数的单调区间显然只由sinx函数决定,所以可以令-pi/2+2kpi