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IAL 2023 May Q9

A Level / Edexcel / P2

IAL 2023 May Paper · Question 9

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(a) Show that

3cos⁡θ (tan⁡θsin⁡θ+3)=11−5cos⁡θ3\cos\theta\,(\tan\theta\sin\theta + 3) = 11 - 5\cos\theta

may be written as

3cos⁡2θ−14cos⁡θ+8=03\cos^2\theta - 14\cos\theta + 8 = 0

(b) Hence solve, for 0°<x<360°0° < x < 360°,

3cos⁡2x (tan⁡2xsin⁡2x+3)=11−5cos⁡2x3\cos 2x\,(\tan 2x\sin 2x + 3) = 11 - 5\cos 2x

giving your answers to one decimal place.

(7)
题目中文翻译

本题必须展示所有解题过程，不能仅依赖计算器。

(a) 证明 3cos⁡θ(tan⁡θsin⁡θ+3)=11−5cos⁡θ3\cos\theta(\tan\theta\sin\theta+3)=11-5\cos\theta 可化为 3cos⁡2θ−14cos⁡θ+8=03\cos^2\theta-14\cos\theta+8=0

(b) 由此解方程 3cos⁡2x(tan⁡2xsin⁡2x+3)=11−5cos⁡2x3\cos 2x(\tan 2x\sin 2x+3)=11-5\cos 2x（0°<x<360°0°<x<360°），答案保留1位小数。

（7分）

解答

解法一

思路

展开左边，用 tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta} 化简，再用 sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta 统一为 cos⁡θ\cos\theta。令 θ=2x\theta=2x，解二次方程得 cos⁡θ\cos\theta，注意 0°<2x<720°0°<2x<720°。

答题过程

(a) Expand the left side:

3\cos\theta(\tan\theta\sin\theta + 3) =&\, 3\cos\theta \cdot \frac{\sin\theta}{\cos\theta} \cdot \sin\theta + 9\cos\theta \\[4mm] =&\, 3\sin^2\theta + 9\cos\theta \end{align*}$$ Using $\sin^2\theta = 1 - \cos^2\theta$: $$\begin{align*} 3(1 - \cos^2\theta) + 9\cos\theta =&\, 11 - 5\cos\theta \\[4mm] 3 - 3\cos^2\theta + 9\cos\theta =&\, 11 - 5\cos\theta \\[4mm] -3\cos^2\theta + 14\cos\theta - 8 =&\, 0 \\[4mm] 3\cos^2\theta - 14\cos\theta + 8 =&\, 0 \end{align*}$$ $$\boxed{3\cos^2\theta - 14\cos\theta + 8 = 0}$$ **(b)** Let $\theta = 2x$. From part (a): $$(3\cos\theta - 2)(\cos\theta - 4) = 0$$ Since $|\cos\theta| \leqslant 1$, $\cos\theta = 4$ is impossible. $$\cos\theta = \frac{2}{3}$$ $$\theta = \cos^{-1}\!\left(\frac{2}{3}\right) \approx 48.2°$$ For $0° < x < 360°$, we have $0° < 2x < 720°$, so: $$2x = 48.2°, \quad 360° - 48.2° = 311.8°, \quad 360° + 48.2° = 408.2°, \quad 720° - 48.2° = 671.8°$$ $$\boxed{x = 24.1°, \quad 155.9°, \quad 204.1°, \quad 335.9°}$$