Question

这题求解！！！

Answer 1

(1) AC = (2sinθ, cosθ) - (1, 0) = (2sinθ - 1, cosθ)
BC = (2sinθ, cosθ) - (0, 1) = (2sinθ, cosθ - 1)
∵∣AC∣= ∣BC∣
∴√[(2sinθ - 1)² + cos²θ] = √[2sin²θ + (cosθ - 1)²]
两边平方, 化简: 4sin²θ - 4sinθ +1 + cos²θ = 4sin²θ + cos²θ - 2cosθ + 1
- 4sinθ = - 2cosθ (cosθ ≠ 0)
tanθ = 1/2

(2) 已知条件是(OA + 2OB) • OC = 1吗??
若是这样的话, 解法如下:
[(1, 0) + 2(0, 1)] • (2sinθ, cosθ) = 1
(1, 2) • (2sinθ, cosθ) = 1
2sinθ + 2cosθ = 1
sinθ + cosθ = 1/2
两边平方, 得: (sinθ + cosθ)² = (1/2)² (注意, 此技巧经常会用到:(sinθ + cosθ)²会得到sinθcosθ, 或 sin2θ的条件)
sin²θ + 2sinθcosθ + cos²θ = 1/4
1 + 2sinθcosθ = 1/4
1 + sin2θ = 1/4
∴ sin2θ = -3/4