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IAL 2021 Oct Q4

A Level / Edexcel / P3

IAL 2021 Oct Paper · Question 4

题目

Problem

In this question you should show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

(a) Show that the equation

2sin⁡(θ−30∘)=5cos⁡θ2\sin(\theta-30^\circ)=5\cos\theta

can be written in the form

tan⁡θ=23\tan\theta=2\sqrt{3}
(4)

(b) Hence, or otherwise, solve for 0≤x≤360∘0\le x\le 360^\circ

2sin⁡(x−10∘)=5cos⁡(x+20∘)2\sin(x-10^\circ)=5\cos(x+20^\circ)

giving your answers to one decimal place.

(3)
题目中文翻译

本题中你应写出详细的推理过程。

不接受完全依赖计算器技术的解法。

(a) 证明方程

2sin⁡(θ−30∘)=5cos⁡θ2\sin(\theta-30^\circ)=5\cos\theta

可以写成

tan⁡θ=23\tan\theta=2\sqrt{3}

的形式。

(b) 由此，或用其他方法，在 0≤x≤360∘0\le x\le 360^\circ 内解方程

2sin⁡(x−10∘)=5cos⁡(x+20∘)2\sin(x-10^\circ)=5\cos(x+20^\circ)

答案精确到小数点后 1 位。

解答

(a)

解法一

思路

展开

展开 sin⁡(θ−30∘)\sin(\theta-30^\circ)，代入 30∘30^\circ 的准确三角函数值，再除以 cos⁡θ\cos\theta 化为关于 tan⁡θ\tan\theta 的方程。

答题过程

展开

Starting with

2sin⁡(θ−30∘)=5cos⁡θ,2\sin(\theta-30^\circ)=5\cos\theta,

use the compound-angle formula:

2(sin⁡θcos⁡30∘−cos⁡θsin⁡30∘)=5cos⁡θ.2\big(\sin\theta\cos30^\circ -\cos\theta\sin30^\circ\big) =5\cos\theta.

Since

cos⁡30∘=32andsin⁡30∘=12,\cos30^\circ=\frac{\sqrt{3}}{2} \qquad\text{and}\qquad \sin30^\circ=\frac{1}{2},

we obtain

3sin⁡θ−cos⁡θ=5cos⁡θ.\sqrt{3}\sin\theta-\cos\theta =5\cos\theta.

Therefore

3sin⁡θ=6cos⁡θ.\sqrt{3}\sin\theta=6\cos\theta.

If cos⁡θ=0\cos\theta=0, this equation would also give sin⁡θ=0\sin\theta=0, which is impossible. Hence cos⁡θ≠0\cos\theta\neq0.

Dividing by cos⁡θ\cos\theta gives

3tan⁡θ=6,\sqrt{3}\tan\theta=6,

and hence

tan⁡θ=23,\boxed{\tan\theta=2\sqrt{3}},

as required.

(b)

解法一

思路

展开

令 θ=x+20∘\theta=x+20^\circ，原方程便完全变成 (a) 的形式。然后使用 (a) 的结论，并把角度换回 xx。

答题过程

展开

Let

θ=x+20∘.\theta=x+20^\circ.

Then

x−10∘=θ−30∘,x-10^\circ=\theta-30^\circ,

so the equation becomes

2sin⁡(θ−30∘)=5cos⁡θ.2\sin(\theta-30^\circ)=5\cos\theta.

Hence, from part (a),

tan⁡θ=23.\tan\theta=2\sqrt{3}.

Since 0≤x≤360∘0\leq x\leq360^\circ, we have

20∘≤θ≤380∘.20^\circ\leq\theta\leq380^\circ.

Therefore

θ=73.897…∘orθ=253.897…∘.\theta=73.897\ldots^\circ \quad\text{or}\quad \theta=253.897\ldots^\circ.

As x=θ−20∘x=\theta-20^\circ,

x=53.897…∘orx=233.897…∘.x=53.897\ldots^\circ \quad\text{or}\quad x=233.897\ldots^\circ.

Thus, to one decimal place,

x=53.9∘, 233.9∘.\boxed{x=53.9^\circ,\ 233.9^\circ}.

解法二

思路

展开

官方评分资料也允许不使用 (a)：直接展开原方程两边的复合角，把 sin⁡x\sin x 项与 cos⁡x\cos x 项分别收集，再求 tan⁡x\tan x。

答题过程

展开

Expand both sides of

2sin⁡(x−10∘)=5cos⁡(x+20∘):2\sin(x-10^\circ)=5\cos(x+20^\circ): 2sin⁡xcos⁡10∘−2cos⁡xsin⁡10∘= 5cos⁡xcos⁡20∘ −5sin⁡xsin⁡20∘.\begin{align*} 2\sin x\cos10^\circ -2\cos x\sin10^\circ =&\,5\cos x\cos20^\circ \\ &\,-5\sin x\sin20^\circ. \end{align*}

Collecting the sin⁡x\sin x and cos⁡x\cos x terms gives

 (2cos⁡10∘+5sin⁡20∘)sin⁡x= (5cos⁡20∘+2sin⁡10∘)cos⁡x.\begin{align*} &\,\big(2\cos10^\circ+5\sin20^\circ\big)\sin x \\ =&\,\big(5\cos20^\circ+2\sin10^\circ\big)\cos x. \end{align*}

The coefficient on the left is non-zero, and this equation has no solution with cos⁡x=0\cos x=0. Therefore

tan⁡x=5cos⁡20∘+2sin⁡10∘2cos⁡10∘+5sin⁡20∘.\tan x =\frac{5\cos20^\circ+2\sin10^\circ} {2\cos10^\circ+5\sin20^\circ}.

For 0≤x≤360∘0\leq x\leq360^\circ,

x=53.897…∘orx=233.897…∘.x=53.897\ldots^\circ \quad\text{or}\quad x=233.897\ldots^\circ.

Hence, to one decimal place,

x=53.9∘, 233.9∘.\boxed{x=53.9^\circ,\ 233.9^\circ}.