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CIE 9709 2026 June Paper 35 Q9

A Level / CIE / P3

CIE 9709 2026 June Paper 35 Paper · Question 9

题目

Problem

(a) Express 5cos⁡θ−3sin⁡θ5\cos\theta-3\sin\theta in the form Rcos⁡(θ+α)R\cos(\theta+\alpha), where R>0R>0 and 0<α<12π0<\alpha<\frac12\pi. State the exact value of RR, and give the value of α\alpha in the form tan⁡−1(p)\tan^{-1}(p), where pp is a rational number.

(3)

(b) Hence, show that, when α\alpha has the value found in 9(a),

∫0α68(5cos⁡θ−3sin⁡θ)2 dθ=5120.\int_0^\alpha\frac{68}{(5\cos\theta-3\sin\theta)^2}\,\mathrm{d}\theta=\frac{51}{20}.
(4)
题目中文翻译

(a) 将 5cos⁡θ−3sin⁡θ5\cos\theta-3\sin\theta 表示为 Rcos⁡(θ+α)R\cos(\theta+\alpha) 的形式，其中 R>0R>0 且 0<α<12π0<\alpha<\frac12\pi。写出 RR 的精确值，并将 α\alpha 的值写成 tan⁡−1(p)\tan^{-1}(p) 的形式，其中 pp 为有理数。

(3)

(b) 由此证明：当 α\alpha 取 9(a) 中求得的值时，

∫0α68(5cos⁡θ−3sin⁡θ)2 dθ=5120.\int_0^\alpha\frac{68}{(5\cos\theta-3\sin\theta)^2}\,\mathrm{d}\theta=\frac{51}{20}.
(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

展开

展开 Rcos⁡(θ+α)R\cos(\theta+\alpha)，比较余弦和正弦的系数。由两系数的平方和求 RR。

答题过程

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Expanding gives Rcos⁡α=5R\cos\alpha=5 and Rsin⁡α=3R\sin\alpha=3. Thus

R2= 52+32=34,R= 34,tan⁡α= 35,α= arctan⁡(3/5),\begin{align*} R^2=&\,5^2+3^2=34,\\[4mm] R=&\,\sqrt{34},\\[4mm] \tan\alpha=&\,\frac35,\\[4mm] \alpha=&\,\arctan(3/5), \end{align*}

with α\alpha in the specified first quadrant. Hence

5cos⁡θ−3sin⁡θ=34cos⁡(θ+α).5\cos\theta-3\sin\theta=\sqrt{34}\cos(\theta+\alpha).

(b)

解法一

思路

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利用 (a) 把被积式改写成正割平方。积分后出现 tan⁡2α\tan2\alpha，用二倍角公式写成分数计算。

答题过程

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Using part (a), the integral is

∫0α2sec⁡2(θ+α) dθ= [2tan⁡(θ+α)]0α= 2(tan⁡2α−tan⁡α).\begin{align*} \int_0^\alpha2\sec^2(\theta+\alpha)\,\mathrm{d}\theta =&\,\bigg[2\tan(\theta+\alpha)\bigg]_0^\alpha\\[4mm] =&\,2(\tan2\alpha-\tan\alpha). \end{align*}

Here tan⁡α=3/5\tan\alpha=3/5 and

tan⁡2α=2(3/5)1−(3/5)2=158.\tan2\alpha=\frac{2(3/5)}{1-(3/5)^2}=\frac{15}{8}.

Also 2α<π/22\alpha<\pi/2, so the integrand is defined throughout the interval. The value is

2(158−35)=5120.2\bigg(\frac{15}{8}-\frac35\bigg)=\frac{51}{20}.