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P2 2025 Jan Q10

A Level / Edexcel / P2

IAL 2025 Jan Paper · Question 10

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

 cos⁡θ(3tan⁡θ+2tan⁡θ)≡ sin⁡θ+2sin⁡θ, θ≠nπ2.\begin{align*} &\,\cos\theta\bigg(3\tan\theta+\frac{2}{\tan\theta}\bigg)\\ \equiv&\,\sin\theta+\frac{2}{\sin\theta},\\ &\,\theta\ne\frac{n\pi}{2}. \end{align*}
(4)

(b) Hence solve, for 0<x<2π0<x<2\pi, the equation

 cos⁡x(3tan⁡x+2tan⁡x)= 4sin⁡x−5.\begin{align*} &\,\cos x\bigg(3\tan x+\frac{2}{\tan x}\bigg)\\ =&\,4\sin x-5. \end{align*}

giving your answers to 3 significant figures.

(4)
题目中文翻译

本题必须展示所有计算步骤。完全依赖计算器技术的解答不可接受。

(a) 证明

 cos⁡θ(3tan⁡θ+2tan⁡θ)≡ sin⁡θ+2sin⁡θ, θ≠nπ2.\begin{align*} &\,\cos\theta\bigg(3\tan\theta+\frac{2}{\tan\theta}\bigg)\\ \equiv&\,\sin\theta+\frac{2}{\sin\theta},\\ &\,\theta\ne\frac{n\pi}{2}. \end{align*}

（4分）

(b) 由此，在 0<x<2π0<x<2\pi 内解方程

 cos⁡x(3tan⁡x+2tan⁡x)= 4sin⁡x−5, \begin{align*} &\,\cos x\bigg(3\tan x+\frac{2}{\tan x}\bigg)\\ =&\,4\sin x-5, \end{align*}

答案保留3位有效数字。（4分）

解答

(a)

解法一

思路

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把 tan⁡θ\tan\theta 写成 sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}。目标式只含 sin⁡θ\sin\theta，所以还要用 cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta。

答题过程

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The given restriction ensures sin⁡θ≠0\sin\theta\ne0 and cos⁡θ≠0\cos\theta\ne0, so all the following divisions are valid. Starting from the left-hand side,

 cos⁡θ(3tan⁡θ+2tan⁡θ)= 3cos⁡θtan⁡θ+2cos⁡θtan⁡θ= 3sin⁡θ+2cos⁡2θsin⁡θ.\begin{align*} &\,\cos\theta\bigg(3\tan\theta+\frac{2}{\tan\theta}\bigg)\\ =&\,3\cos\theta\tan\theta +\frac{2\cos\theta}{\tan\theta}\\ =&\,3\sin\theta+\frac{2\cos^2\theta}{\sin\theta}. \end{align*}

Using cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta,

 3sin⁡θ+2cos⁡2θsin⁡θ= 3sin⁡θ+2(1−sin⁡2θ)sin⁡θ= 3sin⁡θ+2sin⁡θ −2sin⁡θ= sin⁡θ+2sin⁡θ.\begin{align*} &\,3\sin\theta+\frac{2\cos^2\theta}{\sin\theta}\\ =&\,3\sin\theta+\frac{2(1-\sin^2\theta)}{\sin\theta}\\ =&\,3\sin\theta+\frac{2}{\sin\theta}\\ &\,\hspace{2pt}-2\sin\theta\\ =&\,\sin\theta+\frac{2}{\sin\theta}. \end{align*}

Therefore,

 cos⁡θ(3tan⁡θ+2tan⁡θ)≡ sin⁡θ+2sin⁡θ.\begin{align*} &\,\cos\theta\bigg(3\tan\theta+\frac{2}{\tan\theta}\bigg)\\ \equiv&\,\sin\theta+\frac{2}{\sin\theta}. \end{align*}

(b)

解法一

思路

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用第 (a) 题把左边换成 sin⁡x+2sin⁡x\sin x+\frac{2}{\sin x}，然后令 s=sin⁡xs=\sin x 解二次方程。最后要回到 0<x<2π0<x<2\pi 找所有角。

答题过程

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Using part (a), the equation becomes

sin⁡x+2sin⁡x= 4sin⁡x−5.\begin{align*} \sin x+\frac{2}{\sin x}=&\,4\sin x-5. \end{align*}

The original equation requires sin⁡x≠0\sin x\ne0 and cos⁡x≠0\cos x\ne0. Multiply by sin⁡x\sin x:

sin⁡2x+2= 4sin⁡2x−5sin⁡x3sin⁡2x−5sin⁡x−2= 0.\begin{align*} \sin^2x+2=&\,4\sin^2x-5\sin x\\ 3\sin^2x-5\sin x-2=&\,0. \end{align*}

Let s=sin⁡xs=\sin x. Then

3s2−5s−2= 0(3s+1)(s−2)= 0.\begin{align*} 3s^2-5s-2=&\,0\\ (3s+1)(s-2)=&\,0. \end{align*}

So

s=2ors=−13.\begin{align*} s=2\quad\text{or}\quad s=-\frac13. \end{align*}

Since sin⁡x=2\sin x=2 is impossible,

sin⁡x=−13.\begin{align*} \sin x=-\frac13. \end{align*}

For 0<x<2π0<x<2\pi, this gives

x= π+sin⁡−1(13),x= 2π−sin⁡−1(13).\begin{align*} x=&\,\pi+\sin^{-1}\bigg(\frac13\bigg),\\ x=&\,2\pi-\sin^{-1}\bigg(\frac13\bigg). \end{align*}

Therefore,

x=3.48,5.94\begin{align*} x=3.48,\quad 5.94 \end{align*}

to 3 significant figures.