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P3 2022 Oct Q9

A Level / Edexcel / P3

IAL 2022 Oct 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.

Given that cos⁡2θ−sin⁡3θ≠0\cos 2\theta-\sin 3\theta\ne 0

(a) prove that

cos⁡2θcos⁡2θ−sin⁡3θ≡1+sin⁡θ1−2sin⁡θ−4sin⁡2θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}\equiv \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta}
(4)

(b) Hence solve, for 0<θ≤360∘0<\theta\le 360^\circ,

cos⁡2θcos⁡2θ−sin⁡3θ=2cosec⁡θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}=2\cosec\theta

Give your answers to one decimal place.

(5)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

已知 cos⁡2θ−sin⁡3θ≠0\cos 2\theta-\sin 3\theta\ne 0。

(a) 证明

cos⁡2θcos⁡2θ−sin⁡3θ≡1+sin⁡θ1−2sin⁡θ−4sin⁡2θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}\equiv \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta}

(b) 由此在 0<θ≤360∘0<\theta\le 360^\circ 内解方程

cos⁡2θcos⁡2θ−sin⁡3θ=2cosec⁡θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}=2\cosec\theta

答案精确到小数点后 1 位。

解答

(a)

解法一

思路

展开

把 sin⁡3θ\sin 3\theta 展开，再用二倍角与平方和恒等式把分母全部化为 sin⁡θ\sin\theta。分子、分母都出现因子 1−sin⁡θ1-\sin\theta，约去后得到目标式。

答题过程

展开

Using

sin⁡3θ=sin⁡2θcos⁡θ+cos⁡2θsin⁡θ,\sin 3\theta =\sin 2\theta\cos\theta +\cos 2\theta\sin\theta,

the denominator becomes

 cos⁡2θ−sin⁡3θ= 1−2sin⁡2θ−2sin⁡θcos⁡2θ −sin⁡θ(1−2sin⁡2θ)= 1−3sin⁡θ−2sin⁡2θ+4sin⁡3θ= (1−sin⁡θ)(1−2sin⁡θ−4sin⁡2θ)\begin{align*} &\,\cos 2\theta-\sin 3\theta \\ =&\,1-2\sin^2\theta -2\sin\theta\cos^2\theta \\ &\,-\sin\theta(1-2\sin^2\theta) \\ =&\,1-3\sin\theta-2\sin^2\theta +4\sin^3\theta \\ =&\,(1-\sin\theta) (1-2\sin\theta-4\sin^2\theta) \end{align*}

Also,

cos⁡2θ=(1−sin⁡θ)(1+sin⁡θ)\cos^2\theta=(1-\sin\theta)(1+\sin\theta)

Therefore

cos⁡2θcos⁡2θ−sin⁡3θ≡1+sin⁡θ1−2sin⁡θ−4sin⁡2θ\boxed{ \frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta} \equiv \frac{1+\sin\theta} {1-2\sin\theta-4\sin^2\theta} }

as required.

(b)

解法一

思路

展开

按照 Hence 的要求使用 (a) 的恒等式，把方程化为关于 sin⁡θ\sin\theta 的二次方程；求出两个正弦值后，分别在 0∘<θ≤360∘0^\circ<\theta\leq360^\circ 内找全解。

答题过程

展开

Using part (a),

1+sin⁡θ1−2sin⁡θ−4sin⁡2θ=2sin⁡θ\frac{1+\sin\theta} {1-2\sin\theta-4\sin^2\theta} =\frac{2}{\sin\theta}

Cross-multiplying and simplifying,

9sin⁡2θ+5sin⁡θ−2=09\sin^2\theta+5\sin\theta-2=0

Therefore

sin⁡θ=−5±9718\sin\theta=\frac{-5\pm\sqrt{97}}{18}

so

sin⁡θ=0.2693…orsin⁡θ=−0.8249…\sin\theta=0.2693\ldots \quad\text{or}\quad \sin\theta=-0.8249\ldots

For 0∘<θ≤360∘0^\circ<\theta\leq360^\circ, the solutions are

θ=15.6∘, 164.4∘, 235.6∘, 304.4∘\boxed{ \theta=15.6^\circ,\ 164.4^\circ, \ 235.6^\circ,\ 304.4^\circ }