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CIE 9709 2022 June Paper 11 Q4

A Level / CIE / P1

CIE 9709 2022 June Paper 11 Paper · Question 4

题目

Problem

(a) Prove the identity

sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ≡−tan⁡2θ(1+sin⁡2θ).\frac{\sin^3\theta}{\sin\theta-1}-\frac{\sin^2\theta}{1+\sin\theta}\equiv-\tan^2\theta(1+\sin^2\theta).
(4)

(b) Hence solve the equation

sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ=tan⁡2θ(1−sin⁡2θ)\frac{\sin^3\theta}{\sin\theta-1}-\frac{\sin^2\theta}{1+\sin\theta}=\tan^2\theta(1-\sin^2\theta)

for 0<θ<2π0<\theta<2\pi.

(2)
题目中文翻译

(a) 证明恒等式

sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ≡−tan⁡2θ(1+sin⁡2θ).\frac{\sin^3\theta}{\sin\theta-1}-\frac{\sin^2\theta}{1+\sin\theta}\equiv-\tan^2\theta(1+\sin^2\theta).
(4)

(b) 由此解方程

sin⁡3θsin⁡θ−1−sin⁡2θ1+sin⁡θ=tan⁡2θ(1−sin⁡2θ),\frac{\sin^3\theta}{\sin\theta-1}-\frac{\sin^2\theta}{1+\sin\theta}=\tan^2\theta(1-\sin^2\theta),

其中 0<θ<2π0<\theta<2\pi。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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把正弦暂记为 ss。通分后分子的一次项相消，再把分母写成负的余弦平方。

答题过程

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With s=sin⁡θs=\sin\theta, on the common domain the left-hand side becomes

 s3(1+s)−s2(s−1)(s−1)(1+s)= s2(1+s2)s2−1= −s2(1+s2)cos⁡2θ= −tan⁡2θ(1+sin⁡2θ).\begin{align*} &\,\frac{s^3(1+s)-s^2(s-1)}{(s-1)(1+s)}\\ =&\,\frac{s^2(1+s^2)}{s^2-1}\\ =&\,-\frac{s^2(1+s^2)}{\cos^2\theta}\\ =&\,-\tan^2\theta(1+\sin^2\theta). \end{align*}

解法二

思路

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也可从右边出发，将分母分解，再把分子拆成与两个因式相配的两项。

答题过程

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Write s=sin⁡θs=\sin\theta. The right-hand side equals s2(1+s2)/((s−1)(1+s))s^2(1+s^2)/((s-1)(1+s)) on the common domain. Since s2(1+s2)=s3(1+s)−s2(s−1)s^2(1+s^2)=s^3(1+s)-s^2(s-1), splitting this fraction gives s3/(s−1)−s2/(1+s)s^3/(s-1)-s^2/(1+s) as required.

(b)

解法一

思路

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用 (a) 替换左边。移项时含正弦平方的两项抵消，只剩正切平方。

答题过程

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By (a),

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

so 2tan⁡2θ=02\tan^2\theta=0. On 0<θ<2π0<\theta<2\pi, this gives only θ=π\theta=\pi. It is in the original domain and satisfies the equation.