题目

若直线l:xcosθ+ysinθ=cos^2θ-sin^2θ与圆C:x^2+y^2=1/4有公共点,则θ的取值范围是?

提问时间：2020-12-28

答案

圆C圆心为(0,0),半径为1/2
直线l的方程:xcosθ+ysinθ - cos²θ+sin²θ = 0
圆心与直线的距离d = | - cos²θ+sin²θ |/√(cos²θ+sin²θ) = |sin²θ-cos²θ|
= |[1-cos(2θ)]/2 - [1 + cos(2θ)]/2|
= |-cos(2θ)| = |cos(2θ)|
二者有公共点,则d = |cos(2θ)| ≤ 1/2
-1/2 ≤ cos(2θ) ≤ 1/2
π/3 + 2kπ ≤ 2θ ≤ 2π/3 + 2kπ (k为整数)
π/6 + kπ ≤ θ ≤ π/3 + kπ (k为整数)