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CIE 9709 2022 March Paper 12 Q7

A Level / CIE / P1

CIE 9709 2022 March Paper 12 Paper · Question 7

题目

Problem

(a) Show that

sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ≡45cos⁡2θ−4.\frac{\sin\theta+2\cos\theta}{\cos\theta-2\sin\theta}-\frac{\sin\theta-2\cos\theta}{\cos\theta+2\sin\theta}\equiv\frac4{5\cos^2\theta-4}.
(4)

(b) Hence solve the equation

sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ=5\frac{\sin\theta+2\cos\theta}{\cos\theta-2\sin\theta}-\frac{\sin\theta-2\cos\theta}{\cos\theta+2\sin\theta}=5

for 0∘<θ<180∘0^\circ<\theta<180^\circ.

(3)
题目中文翻译

(a) 证明

sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ≡45cos⁡2θ−4.\frac{\sin\theta+2\cos\theta}{\cos\theta-2\sin\theta}-\frac{\sin\theta-2\cos\theta}{\cos\theta+2\sin\theta}\equiv\frac4{5\cos^2\theta-4}.
(4)

(b) 由此解方程

sin⁡θ+2cos⁡θcos⁡θ−2sin⁡θ−sin⁡θ−2cos⁡θcos⁡θ+2sin⁡θ=5,\frac{\sin\theta+2\cos\theta}{\cos\theta-2\sin\theta}-\frac{\sin\theta-2\cos\theta}{\cos\theta+2\sin\theta}=5,

其中 0∘<θ<180∘0^\circ<\theta<180^\circ。

(3)
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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令 s=sin⁡θs=\sin\theta、c=cos⁡θc=\cos\theta。通分后，分子用平方关系化为常数，分母用平方差化简。

答题过程

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Let s=sin⁡θs=\sin\theta and c=cos⁡θc=\cos\theta. On the common domain, the numerator after taking a common denominator is

 (s+2c)(c+2s)−(s−2c)(c−2s)= 5sc+2s2+2c2−(5sc−2s2−2c2)= 4s2+4c2=4.\begin{align*} &\,(s+2c)(c+2s)-(s-2c)(c-2s)\\ =&\,5sc+2s^2+2c^2-\Big(5sc-2s^2-2c^2\Big)\\ =&\,4s^2+4c^2=4. \end{align*}

The denominator is (c−2s)(c+2s)=c2−4s2=5c2−4(c-2s)(c+2s)=c^2-4s^2=5c^2-4. This gives the stated identity.

(b)

解法一

思路

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用 (a) 化成余弦平方方程。开平方后分别求正、负余弦对应的角。

答题过程

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From 4/(5cos⁡2θ−4)=54/(5\cos^2\theta-4)=5, we obtain 25cos⁡2θ=2425\cos^2\theta=24. Thus cos⁡θ=±26/5\cos\theta=\pm2\sqrt6/5. In the stated interval the solutions are 11.5∘11.5^\circ and 168.5∘168.5^\circ. At these angles the common denominator is 4/54/5, so the original fractions are defined.