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CIE 9709 2023 June Paper 12 Q7

A Level / CIE / P1

CIE 9709 2023 June Paper 12 Paper · Question 7

题目

Problem

(a) (i) By first expanding (cos⁡θ+sin⁡θ)2(\cos\theta + \sin\theta)^2, find the three solutions of the equation

(cos⁡θ+sin⁡θ)2=1(\cos\theta + \sin\theta)^2 = 1

for 0≤θ≤π0 \leq \theta \leq \pi.

[3]

(ii) Hence verify that the only solutions of the equation cos⁡θ+sin⁡θ=1\cos\theta + \sin\theta = 1 for 0≤θ≤π0 \leq \theta \leq \pi are 00 and 12π\frac12\pi.

[2]

(b) Prove the identity

sin⁡θcos⁡θ+sin⁡θ+1−cos⁡θcos⁡θ−sin⁡θ=cos⁡θ+sin⁡θ−11−2sin⁡2θ.\frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{1 - \cos\theta}{\cos\theta - \sin\theta} = \frac{\cos\theta + \sin\theta - 1}{1 - 2\sin^2\theta}.
[3]

(c) Using the results of (a)(ii) and (b), solve the equation

sin⁡θcos⁡θ+sin⁡θ+1−cos⁡θcos⁡θ−sin⁡θ=2(cos⁡θ+sin⁡θ−1)\frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{1 - \cos\theta}{\cos\theta - \sin\theta} = 2(\cos\theta + \sin\theta - 1)

for 0≤θ≤π0 \leq \theta \leq \pi.

[3]
题目中文翻译

(a) (i) 先展开 (cos⁡θ+sin⁡θ)2(\cos\theta + \sin\theta)^2，求方程

(cos⁡θ+sin⁡θ)2=1(\cos\theta + \sin\theta)^2 = 1

在 0≤θ≤π0 \leq \theta \leq \pi 中的三个解。

[3]

(ii) 由此验证方程 cos⁡θ+sin⁡θ=1\cos\theta + \sin\theta = 1 在 0≤θ≤π0 \leq \theta \leq \pi 中的唯一解为 00 和 12π\frac12\pi。

[2]

(b) 证明恒等式

sin⁡θcos⁡θ+sin⁡θ+1−cos⁡θcos⁡θ−sin⁡θ=cos⁡θ+sin⁡θ−11−2sin⁡2θ.\frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{1 - \cos\theta}{\cos\theta - \sin\theta} = \frac{\cos\theta + \sin\theta - 1}{1 - 2\sin^2\theta}.
[3]

(c) 利用 (a)(ii) 和 (b) 的结果，解方程

sin⁡θcos⁡θ+sin⁡θ+1−cos⁡θcos⁡θ−sin⁡θ=2(cos⁡θ+sin⁡θ−1)\frac{\sin\theta}{\cos\theta + \sin\theta} + \frac{1 - \cos\theta}{\cos\theta - \sin\theta} = 2(\cos\theta + \sin\theta - 1)

其中 0≤θ≤π0 \leq \theta \leq \pi。

[3]

解答