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CIE 9709 2023 March Paper 12 Q7

A Level / CIE / P1

CIE 9709 2023 March Paper 12 Paper · Question 7

题目

Problem

(a) By first obtaining a quadratic equation in cos⁡θ\cos\theta, solve the equation

tan⁡θsin⁡θ=1\tan\theta\sin\theta=1

for 0∘<θ<360∘0^\circ<\theta<360^\circ.

(5)

(b) Show that

tan⁡θsin⁡θ−sin⁡θtan⁡θ≡tan⁡θsin⁡θ.\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}\equiv\tan\theta\sin\theta.
(3)
题目中文翻译

(a) 先列出关于 cos⁡θ\cos\theta 的二次方程，再解方程

tan⁡θsin⁡θ=1,\tan\theta\sin\theta=1,

其中 0∘<θ<360∘0^\circ<\theta<360^\circ。

(5)

(b) 证明

tan⁡θsin⁡θ−sin⁡θtan⁡θ≡tan⁡θsin⁡θ.\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}\equiv\tan\theta\sin\theta.
(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

展开

用正切的定义消去分母，再用平方关系得到余弦的二次式。先筛选余弦值，再找角度。

答题过程

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For cos⁡θ≠0\cos\theta\ne0, the equation is sin⁡2θ=cos⁡θ\sin^2\theta=\cos\theta, so

cos⁡2θ+cos⁡θ−1=0.\cos^2\theta+\cos\theta-1=0.

The quadratic formula gives

cos⁡θ=−1±12−4(1)(−1)2=−1±52.\cos\theta=\frac{-1\pm\sqrt{1^2-4(1)(-1)}}{2}=\frac{-1\pm\sqrt5}{2}.

Only (5−1)/2(\sqrt5-1)/2 lies in [−1,1][-1,1]. The corresponding angles in the stated interval are 51.8∘51.8^\circ and 308.2∘308.2^\circ, and both satisfy the original equation.

(b)

解法一

思路

展开

先用正弦和余弦改写正切。两项化简后通分，再用平方关系。

答题过程

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Where the original fractions are defined,

 tan⁡θsin⁡θ−sin⁡θtan⁡θ= 1cos⁡θ−cos⁡θ= 1−cos⁡2θcos⁡θ= sin⁡2θcos⁡θ= tan⁡θsin⁡θ.\begin{align*} &\,\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}\\ =&\,\frac1{\cos\theta}-\cos\theta\\ =&\,\frac{1-\cos^2\theta}{\cos\theta}\\ =&\,\frac{\sin^2\theta}{\cos\theta}\\ =&\,\tan\theta\sin\theta. \end{align*}