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P2 2024 Oct Q9

A Level / Edexcel / P2

IAL 2024 Oct Paper · Question 9

题目

Problem

In this question you must show detailed reasoning. Solutions relying entirely on calculator technology are not acceptable.

(a) Show that the equation

2tan⁡θ=3cos⁡θ2\tan\theta=3\cos\theta

can be written as

3sin⁡2θ+2sin⁡θ−3=03\sin^2\theta+2\sin\theta-3=0
(3)

(b) Hence solve, for −π<x<π-\pi<x<\pi, the equation

 2tan⁡(2x+π3)= 3cos⁡(2x+π3)\begin{align*} &\,2\tan\bigg(2x+\frac{\pi}{3}\bigg)\\ =&\,3\cos\bigg(2x+\frac{\pi}{3}\bigg) \end{align*}

giving your answers to 3 significant figures.

(4)
题目中文翻译

本题必须展示详细的推理过程。

完全依赖计算器技术的解答不予接受。

(a) 证明方程

2tan⁡θ=3cos⁡θ2\tan\theta=3\cos\theta

可以写为

3sin⁡2θ+2sin⁡θ−3=03\sin^2\theta+2\sin\theta-3=0

(3)

(b) 由此，在 −π<x<π-\pi<x<\pi 内解以下方程，答案精确到3位有效数字。(4)

 2tan⁡(2x+π3)= 3cos⁡(2x+π3)\begin{align*} &\,2\tan\bigg(2x+\frac{\pi}{3}\bigg)\\ =&\,3\cos\bigg(2x+\frac{\pi}{3}\bigg) \end{align*}

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}，然后用 cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta。

答题过程

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Starting with

2tan⁡θ= 3cos⁡θ,\begin{align*} 2\tan\theta=&\,3\cos\theta, \end{align*}

use tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}:

2⋅sin⁡θcos⁡θ= 3cos⁡θ.\begin{align*} 2\cdot\frac{\sin\theta}{\cos\theta}=&\,3\cos\theta. \end{align*}

The original equation contains tan⁡θ\tan\theta, so cos⁡θ≠0\cos\theta\neq0. Multiply by cos⁡θ\cos\theta:

2sin⁡θ= 3cos⁡2θ.\begin{align*} 2\sin\theta=&\,3\cos^2\theta. \end{align*}

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

2sin⁡θ= 3(1−sin⁡2θ)2sin⁡θ= 3−3sin⁡2θ3sin⁡2θ+2sin⁡θ−3= 0.\begin{align*} 2\sin\theta=&\,3(1-\sin^2\theta)\\ 2\sin\theta=&\,3-3\sin^2\theta\\ 3\sin^2\theta+2\sin\theta-3=&\,0. \end{align*}

This is the required form.

也可逆向证明：由目标二次式得 2sin⁡θ=3(1−sin⁡2θ)=3cos⁡2θ2\sin\theta=3(1-\sin^2\theta)=3\cos^2\theta。在原式的定义域 cos⁡θ≠0\cos\theta\neq0 内，两边除以 cos⁡θ\cos\theta，便得到 2tan⁡θ=3cos⁡θ2\tan\theta=3\cos\theta。

(b)

解法一

思路

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令 θ=2x+π3\theta=2x+\frac{\pi}{3}，直接用 (a) 得到关于 sin⁡θ\sin\theta 的二次方程。然后把所有 θ\theta 的解转回 xx，并筛选 −π<x<π-\pi<x<\pi。

答题过程

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Let

θ= 2x+π3.\begin{align*} \theta=&\,2x+\frac{\pi}{3}. \end{align*}

Using part (a),

3sin⁡2θ+2sin⁡θ−3= 0.\begin{align*} 3\sin^2\theta+2\sin\theta-3=&\,0. \end{align*}

Solving this quadratic,

sin⁡θ= −2±22−4(3)(−3)2(3)= −2±406= −1±103.\begin{align*} \sin\theta =&\,\frac{-2\pm\sqrt{2^2-4(3)(-3)}}{2(3)}\\ =&\,\frac{-2\pm\sqrt{40}}{6}\\ =&\,\frac{-1\pm\sqrt{10}}{3}. \end{align*}

Only

sin⁡θ= −1+103\begin{align*} \sin\theta=&\,\frac{-1+\sqrt{10}}{3} \end{align*}

is possible, since the other root is less than −1-1.

Let

α=arcsin⁡(10−13)=0.804896958651….\alpha=\arcsin\bigg(\frac{\sqrt{10}-1}{3}\bigg) =0.804896958651\ldots.

The given range −π<x<π-\pi<x<\pi becomes

−5π3<θ<7π3.-\frac{5\pi}{3}<\theta<\frac{7\pi}{3}.

The two sine branches are

θ= α+2nπor θ= π−α+2nπ,n∈Z.\begin{align*} \theta=&\,\alpha+2n\pi\\ \text{or }\theta=&\,\pi-\alpha+2n\pi, \qquad n\in\mathbb Z. \end{align*}

The four values in this interval are

θ1= π−α−2πθ2= αθ3= π−αθ4= α+2π.\begin{align*} \theta_1=&\,\pi-\alpha-2\pi\\ \theta_2=&\,\alpha\\ \theta_3=&\,\pi-\alpha\\ \theta_4=&\,\alpha+2\pi. \end{align*}

Using x=θ−π/32x=\dfrac{\theta-\pi/3}{2} for each value,

x1= −2.496843581718…x2= −0.121150296272…x3= 0.644749071870…x4= 3.020442357317….\begin{align*} x_1=&\,-2.496843581718\ldots\\ x_2=&\,-0.121150296272\ldots\\ x_3=&\,0.644749071870\ldots\\ x_4=&\,3.020442357317\ldots. \end{align*}

All four are in −π<x<π-\pi<x<\pi, and none makes the original tangent undefined. Therefore, to 3 significant figures,

x= −2.50, −0.121, 0.645, 3.02.\begin{align*} x=&\,-2.50,\ -0.121,\\ &\,\hspace{2pt}0.645,\ 3.02. \end{align*}