先求B(3,√3)的极坐标;ρ²=3²+(√3)²=12,ρ可取2√3
tanθ=√3/3==>θ=π/6
所以B(2√3,π/6)
再求A(-3,√3)的极坐标;ρ²=3²+(√3)²=12,ρ可取2√3
tanθ=-√3/3,且点在二象限,所以θ=5π/6
所以A(2√3,5π/6)(3,-根号3)呢ρ=2√3,tanθ= - √3/3,θ= - π/6C(2√3,- π/6)注:极坐标的表达是不唯一的,也可心用:C(2√3,7π/6),还可以用负径表示,这有一点难!C(-2√3,5π/6)
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