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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

(a)(i)

解法一

思路

展开

先展开平方。利用平方恒等式后，只剩正弦与余弦乘积为零。

答题过程

展开

Expanding gives cos⁡2θ+2sin⁡θcos⁡θ+sin⁡2θ=1\cos^2\theta+2\sin\theta\cos\theta+\sin^2\theta=1. Thus 2sin⁡θcos⁡θ=02\sin\theta\cos\theta=0. In [0,π][0,\pi], the zeros of sine are 0,π0,\pi, and the zero of cosine is π/2\pi/2. These are the three solutions.

(a)(ii)

解法一

思路

展开

平方方程的解可能使原式等于 −1-1。把 (i) 的三个候选值逐一代回。

答题过程

展开

At θ=0\theta=0 and θ=π/2\theta=\pi/2, the sum cos⁡θ+sin⁡θ\cos\theta+\sin\theta is 11. At θ=π\theta=\pi it is −1-1. Since any solution of the unsquared equation solves its square, the only solutions are 00 and π/2\pi/2.

(b)

解法一

思路

展开

用两个分母的乘积通分。展开分子后，混合项相消，两个平方项再合成 11。

答题过程

展开

Where both denominators are nonzero, the left side has denominator cos⁡2θ−sin⁡2θ\cos^2\theta-\sin^2\theta. Its numerator is

 sin⁡θ(cos⁡θ−sin⁡θ) +(1−cos⁡θ)(cos⁡θ+sin⁡θ)= sin⁡θcos⁡θ−sin⁡2θ+cos⁡θ +sin⁡θ−cos⁡2θ−sin⁡θcos⁡θ= cos⁡θ+sin⁡θ−sin⁡2θ−cos⁡2θ= cos⁡θ+sin⁡θ−1.\begin{align*} &\,\sin\theta(\cos\theta-\sin\theta)\\ &\,+(1-\cos\theta)(\cos\theta+\sin\theta)\\ =&\,\sin\theta\cos\theta-\sin^2\theta+\cos\theta\\ &\,+\sin\theta-\cos^2\theta-\sin\theta\cos\theta\\ =&\,\cos\theta+\sin\theta-\sin^2\theta-\cos^2\theta\\ =&\,\cos\theta+\sin\theta-1. \end{align*}

Since the denominator is 1−2sin⁡2θ1-2\sin^2\theta, the identity follows.

(c)

解法一

思路

展开

用 (b) 化简左边，但不要直接约掉 cos⁡θ+sin⁡θ−1\cos\theta+\sin\theta-1。这个因子为零的情形就是 (a)(ii)。

答题过程

展开

By (b) the equation becomes

(cos⁡θ+sin⁡θ−1)(11−2sin⁡2θ−2)=0.(\cos\theta+\sin\theta-1) \Big(\frac1{1-2\sin^2\theta}-2\Big)=0.

The first factor is zero at 00 and π/2\pi/2 by (a)(ii). The second is zero when 1−2sin⁡2θ=1/21-2\sin^2\theta=1/2, so sin⁡2θ=1/4\sin^2\theta=1/4. Sine is nonnegative on [0,π][0,\pi], giving π/6\pi/6 and 5π/65\pi/6. None of the four angles makes either original denominator zero. The solutions are 00, π/6\pi/6, π/2\pi/2 and 5π/65\pi/6.