sin(α+π/3)+sinα=负2分之根号3 α∈（-π/2,0）求cosα

f(x)=2sin[(ωx+π/3+ωx-π/3)]/2*cos[(ωx+π/3-ωx+π/3)/2]+√3cosωx.
=2sinωx*cos(π/3)+√3cosωx.
=sinωx+√3cosωx.
∴f(x)=2sin(ωx+π/3).
(1) 由题设得：T/2=π/2,∴T=π；
ω=2π/T=2π/π
∴ω=2.
∴f(x)=2sin(2x+π/3).
(2) 将f(x)=2sin(2x+π/3)的图象上的横坐标伸长至原来的2倍,得g(x)=2sin(x+π/3).【原来x的系数÷2即2÷2＝1】
g(x)=2sin(x+π/3)在x∈[0,π/2]区间上的最大值和最小值：
g(0)=2sin(0+π/3)=2sinπ/3=√3
g(π/2)=2sin(π/2+π/3)=2cosπ/3=2*(1/2)=1.
∴函数g(x)在x∈[0,π/2}区间上的最大值和最小值分别是：g(x)max=√3.,g(x)min=1.

f(x)=sin(wx+2π/3)+sin(wx-2π/3)
=sinwxcos(2π/3)+coswxsin(2π/3)+sinwxcos(2π/3)-coswxsin(2π/3)
=sinwxcos(2π/3)+sinwxcos(2π/3)
=2sinwxcos(2π/3)
=-sinwx
∵T=π
∴2π/w=π
w=2
∴函数为f(x)=-sin2x
∴函数的单调增区间为：（π/2+kπ,3π/2+kπ）k∈Z
函数的单调减区间为：（-π/2+kπ,π/2+kπ) k∈Z

sin(α+π/3)+sinα
=[sina/2+√3cosa/2]+sina
=3sina/2+√3cosa/2
=√3[√3sina/2+cosa/2]
=√3sin(a+π/6)=-√3/2
sin(a+π/6)=-1/2
a=-π/3
cosa=cos(π/3)=1/2

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+3sin^2x
=1-2sin^2x+cosx+3sin^2x
=1+sin^2x+cosx
=-cosx^2+cosx+2
=-(cosx-1/2)^2+9/4
当cosx=-1时,
所取得值最小为0

两边合并同类项,然后同除以3

f(B+π/2)=1
即
sin(B+π/2+π/3)+sin(B+π/2-π/3)+√3cos(B+π/2)+1=1
所以
sin(B+5π/6) +sin(B+π/6) -√3sinB=0
展开得到
sinB*cos(5π/6) +cosB*sin(5π/6)+sinB*cos(π/6) +cosB*sin(π/6) -√3sinB=0
显然sin(5π/6)=sin(π/6)=1/2,cos(5π/6)= -cos(π/6)
那么
cosB-√3sinB=0
即tanB=√3 /3
所以B=30度
再由正弦定理,
b/sinB=c/sinC
故1/(1/2)= √3 /sinC,得到sinC=√3/2,所以C=60或120度,
若C为60度,则A为90度,a=√(b^2+c^2)=2,
若C为120度,则A为30度,a=b=1