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CIE 9709 2025 November Paper 32 Q8

A Level / CIE / P3

CIE 9709 2025 Nov Paper 32 Paper · Question 8

题目

Problem

(a) Prove the identity sin⁡4x≡4sin⁡x(2cos⁡3x−cos⁡x)\sin4x\equiv4\sin x(2\cos^3x-\cos x).

(3)

(b) Hence find the exact value of ∫014πcos⁡3xsin⁡4x dx\displaystyle\int_0^{\frac14\pi}\cos^3x\sin4x\,\mathrm{d}x.

(5)
题目中文翻译

(a) 证明恒等式 sin⁡4x≡4sin⁡x(2cos⁡3x−cos⁡x)\sin4x\equiv4\sin x(2\cos^3x-\cos x)。

(3)

(b) 由此求 ∫014πcos⁡3xsin⁡4x dx\displaystyle\int_0^{\frac14\pi}\cos^3x\sin4x\,\mathrm{d}x 的精确值。

(5)
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

连续用两次二倍角公式,再把 cos⁡2x\cos2x 写成仅含余弦的形式。

答题过程

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Using the double-angle identities,

sin⁡4x≡ 2sin⁡2xcos⁡2x≡ 2(2sin⁡xcos⁡x)(2cos⁡2x−1)≡ 4sin⁡x(2cos⁡3x−cos⁡x),\begin{align*} \sin4x \equiv&\,2\sin2x\cos2x\\ \equiv&\,2(2\sin x\cos x)(2\cos^2x-1)\\ \equiv&\,4\sin x(2\cos^3x-\cos x), \end{align*}

as required.

(b)

解法一

思路

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承接(a),让被积函数出现 sin⁡x\sin x 乘余弦的幂,然后用链式法则的逆过程积分。代入端点时注意根号的幂次。

答题过程

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Using the identity from (a),

cos⁡3xsin⁡4x=8sin⁡xcos⁡6x−4sin⁡xcos⁡4x.\cos^3x\sin4x =8\sin x\cos^6x-4\sin x\cos^4x.

With u=cos⁡xu=\cos x and du=−sin⁡x dx\mathrm{d}u=-\sin x\,\mathrm{d}x, the limits become u=1u=1 and u=2/2u=\sqrt2/2. Thus

I=∫2/21(8u6−4u4) du.I=\int_{\sqrt2/2}^{1}(8u^6-4u^4)\,\mathrm{d}u.

Equivalently, integrating in terms of cos⁡x\cos x,

I=[−87cos⁡7x+45cos⁡5x]0π/4.I=\bigg[-\frac87\cos^7x+\frac45\cos^5x\bigg]_0^{\pi/4}.

At x=π/4x=\pi/4,

 −87(22)7+45(22)5= −214+210= 235.\begin{align*} &\,-\frac87\bigg(\frac{\sqrt2}{2}\bigg)^7 +\frac45\bigg(\frac{\sqrt2}{2}\bigg)^5\\[4mm] =&\,-\frac{\sqrt2}{14}+\frac{\sqrt2}{10}\\[4mm] =&\,\frac{\sqrt2}{35}. \end{align*}

At x=0x=0, the bracket is −8/7+4/5=−12/35-8/7+4/5=-12/35. Therefore

I=12+235.\boxed{I=\frac{12+\sqrt2}{35}}.