a向量=(根号3cosx/2,2cosx/2),向量b=(2cosx/2,-sinx/2)函数f(x)=向量a·向量b1）设θ∈[-π/2,π/2],且f（θ）=根号3 +1,求θ的值2）在△ABC中,AB=1,f（C）=根号3 +1,且△ABC的面积为 根号3 /2,求sinA+sinB的值

f(x)=a·b=(sqrt(3)cosx/2,2cosx/2)·(2cosx/2,-sinx/2)=2sqrt(3)cos(x/2)^2-sinx
=sqrt(3)cosx-sinx+sqrt(3)=sqrt(3)-2sin(x-π/3)
1
t表示theta：f(t)=sqrt(3)-2sin(t-π/3)=sqrt(3)+1，故：sin(t-π/3)=-1/2
-π/2≤t≤π/2，故：-5π/6≤t-π/3≤π/6，故：t-π/3=-π/6或-5π/6，即：t=π/6或-π/2
2
f(C)=sqrt(3)-2sin(C-π/3)=sqrt(3)+1，即：sin(C-π/3)=-1/2，C是内角，故：0故：-π/3△ABC的面积：S=(1/2)absinC=ab/4=sqrt(3)/2，即：ab=2sqrt(3)
而：a^2+b^2-2abcosC=c^2=1，即：a^2+b^2=1+2abcosC=7
即：a^2+b^2-2ab=7-4sqrt(3)，即：(a-b)^2=(2-sqrt(3))^2，即：a-b=2-sqrt(3)或sqrt(3)-2
故：a=2，b=sqrt(3)或a=sqrt(3)，b=2，故△ABC是以A或B为直角的直角三角形
故：sinA+sinB=1+sin(π/3)=1+sqrt(3)/2

（1）f（x）=ab=2√3cosx/2*cosx/2-2sinx/2cosx/2=√3（1+cosx）-sinx=2sin（x+2π/3）+√3f（θ）=根号3 +1得到2sin（θ+2π/3）+√3=√3+1 得到sin（θ+2π/3）=1/2而θ∈[-π/2,π/2] 得到θ+2π/3属于[π/6,7π/...