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CIE 9709 2026 June Paper 13 Q3

A Level / CIE / P1

CIE 9709 2026 June Paper 13 Paper · Question 3

题目

Problem

(a) Prove the identity

1cos⁡θ−cos⁡θ≡tan⁡θsin⁡θ.\frac{1}{\cos\theta}-\cos\theta\equiv\tan\theta\sin\theta.
(3)

(b) Hence, solve the equation

2cos⁡θ−2cos⁡θ=tan⁡θ\frac{2}{\cos\theta}-2\cos\theta=\tan\theta

for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

(4)
题目中文翻译

(a) 证明恒等式

1cos⁡θ−cos⁡θ≡tan⁡θsin⁡θ.\frac{1}{\cos\theta}-\cos\theta\equiv\tan\theta\sin\theta.
(3)

(b) 由此，在 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ 内解方程

2cos⁡θ−2cos⁡θ=tan⁡θ.\frac{2}{\cos\theta}-2\cos\theta=\tan\theta.
(4)
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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先通分，再用 1−cos⁡2θ=sin⁡2θ1-\cos^2\theta=\sin^2\theta。把一份正弦与分母组成正切。

答题过程

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For cos⁡θ≠0\cos\theta\ne0,

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

(b)

解法一

思路

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用 (a) 改写左式。移项后提取正切因子，避免漏掉正切为零的角度。

答题过程

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By (a), 2tan⁡θsin⁡θ=tan⁡θ2\tan\theta\sin\theta=\tan\theta, so

tan⁡θ(2sin⁡θ−1)=0.\tan\theta(2\sin\theta-1)=0.

In the stated interval, tan⁡θ=0\tan\theta=0 at 0∘0^\circ, 180∘180^\circ and 360∘360^\circ. Also sin⁡θ=1/2\sin\theta=1/2 at 30∘30^\circ and 150∘150^\circ. All five angles have cos⁡θ≠0\cos\theta\ne0, so all satisfy the original equation.