CalcGospel 國際數學圖譜
返回

P2 2022 Oct Q5

A Level / Edexcel / P2

IAL 2022 Oct Paper · Question 5

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

(3cos⁡θ−tan⁡θ)cos⁡θ=2(3 \cos \theta - \tan \theta)\cos \theta = 2

can be written as

3sin⁡2θ+sin⁡θ−1=03 \sin^2 \theta + \sin \theta - 1 = 0
(3)

(b) Hence solve for −π2⩽x⩽π2-\dfrac{\pi}{2} \leqslant x \leqslant \dfrac{\pi}{2}

(3cos⁡2x−tan⁡2x)cos⁡2x=2(3 \cos 2x - \tan 2x)\cos 2x = 2
(5)
题目中文翻译

本题必须展示详细推理。完全依赖计算器技术的解答不予接受。

(a) 证明方程

(3cos⁡θ−tan⁡θ)cos⁡θ=2(3\cos\theta-\tan\theta)\cos\theta=2

可写成

3sin⁡2θ+sin⁡θ−1=0.3\sin^2\theta+\sin\theta-1=0.

（3 分）

(b) 由此在 −π/2⩽x⩽π/2-\pi/2\leqslant x\leqslant\pi/2 范围内解方程

(3cos⁡2x−tan⁡2x)cos⁡2x=2.(3\cos2x-\tan2x)\cos2x=2.

（5 分）

解答

(a)

解法一

思路

展开

首先，将方程式左边展开：

3cos⁡2θ−tan⁡θcos⁡θ=23\cos^2\theta - \tan\theta\cos\theta = 2

利用正切函数的定义式将 tan⁡θ\tan\theta 转化为正弦和余弦：

tan⁡θ= sin⁡θcos⁡θtan⁡θcos⁡θ= sin⁡θcos⁡θcos⁡θ=sin⁡θ\begin{align*} \tan\theta=&\,\frac{\sin\theta}{\cos\theta}\\ \tan\theta\cos\theta=&\,\frac{\sin\theta}{\cos\theta}\cos\theta = \sin\theta \end{align*}

代回原式可得：

3cos⁡2θ−sin⁡θ=23\cos^2\theta - \sin\theta = 2

由于最终方程只包含正弦函数 sin⁡θ\sin\theta，我们使用毕达哥拉斯三角恒等式（Pythagorean identity）cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta 进行代换：

3(1−sin⁡2θ)−sin⁡θ=23(1 - \sin^2\theta) - \sin\theta = 2

展开整理：

3−3sin⁡2θ−sin⁡θ= 23sin⁡2θ+sin⁡θ−1= 0\begin{align*} 3 - 3\sin^2\theta - \sin\theta=&\,2\\ 3\sin^2\theta + \sin\theta - 1=&\,0 \end{align*}

即可得到要证明的式子。

答题过程

展开

Expand the left-hand side of the given equation:

3cos⁡2θ−tan⁡θcos⁡θ=23\cos^2\theta - \tan\theta\cos\theta = 2

Using the identity tan⁡θ=sin⁡θcos⁡θ\tan\theta = \dfrac{\sin\theta}{\cos\theta} (for cos⁡θ≠0\cos\theta \neq 0):

3cos⁡2θ−(sin⁡θcos⁡θ)cos⁡θ= 23cos⁡2θ−sin⁡θ= 2\begin{align*} 3\cos^2\theta - \Big(\frac{\sin\theta}{\cos\theta}\Big)\cos\theta =&\, 2 \\ 3\cos^2\theta - \sin\theta =&\, 2 \end{align*}

Using the Pythagorean identity cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta:

3(1−sin⁡2θ)−sin⁡θ= 23−3sin⁡2θ−sin⁡θ= 2−3sin⁡2θ−sin⁡θ+1= 0\begin{align*} 3(1 - \sin^2\theta) - \sin\theta =&\, 2 \\ 3 - 3\sin^2\theta - \sin\theta =&\, 2 \\ -3\sin^2\theta - \sin\theta + 1 =&\, 0 \end{align*}

Multiplying the entire equation by −1-1:

3sin⁡2θ+sin⁡θ−1=0(proven)3\sin^2\theta + \sin\theta - 1 = 0 \quad \text{(proven)}

(b)

解法一

思路

展开

由 (a) 将 θ\theta 换成 2x2x，方程变成 3sin⁡22x+sin⁡2x−1=03\sin^2 2x+\sin2x-1=0。用求根公式得到

sin⁡2x=−1±136.\sin2x=\frac{-1\pm\sqrt{13}}6.

因为 −π/2⩽x⩽π/2-\pi/2\leqslant x\leqslant\pi/2，求角时须使用 −π⩽2x⩽π-\pi\leqslant2x\leqslant\pi。

正根的反正弦主值为 α=0.449214996…\alpha=0.449214996\ldots，范围内另一角为 π−α\pi-\alpha。负根的主值为 β=−0.875075474…\beta=-0.875075474\ldots，范围内另一角为 −π−β-\pi-\beta。将四个角都除以 2，才得到 xx。

计算中使用精确根及未舍入的角度，最后把 xx 舍入到三位有效数字。

答题过程

展开

Let θ=2x\theta = 2x. From the conclusion in part (a), the equation can be rewritten as:

3sin⁡22x+sin⁡2x−1=03 \sin^2 2x + \sin 2x - 1 = 0

Given the range for xx is −π2⩽x⩽π2-\dfrac{\pi}{2} \leqslant x \leqslant \dfrac{\pi}{2}, the range for 2x2x is:

−π⩽2x⩽π-\pi \leqslant 2x \leqslant \pi

Solve the quadratic equation in sin⁡2x\sin 2x using the quadratic formula:

sin⁡2x= −1±12−4(3)(−1)2(3)= −1±136\begin{align*} \sin 2x =&\, \frac{-1 \pm \sqrt{1^2 - 4(3)(-1)}}{2(3)} \\ =&\, \frac{-1 \pm \sqrt{13}}{6} \end{align*}

This gives two possible values for sin⁡2x\sin 2x:

  1. sin⁡2x=−1+136≈0.43426\sin 2x = \dfrac{-1 + \sqrt{13}}{6} \approx 0.43426
  2. sin⁡2x=−1−136≈−0.76759\sin 2x = \dfrac{-1 - \sqrt{13}}{6} \approx -0.76759

We find the solutions for 2x2x within the interval [−π,π][-\pi, \pi]:

Case 1: When sin⁡2x≈0.43426\sin 2x \approx 0.43426 (positive root)

The principal value is:

2x=arcsin⁡(−1+136)≈0.449214996 rad2x = \arcsin\Big(\frac{-1+\sqrt{13}}6\Big) \approx 0.449214996\text{ rad}

The other solution in the range [−π,π][-\pi, \pi] is:

2x=π−0.449214996≈2.692377657 rad2x = \pi - 0.449214996 \approx 2.692377657\text{ rad}

Dividing by 22 to find xx:

x= 0.4492149962≈0.225 radx= 2.6923776572≈1.35 rad\begin{align*} x =&\, \frac{0.449214996}{2} \approx 0.225\text{ rad} \\ x =&\, \frac{2.692377657}{2} \approx 1.35\text{ rad} \end{align*}

Case 2: When sin⁡2x≈−0.76759\sin 2x \approx -0.76759 (negative root)

The principal value is:

2x=arcsin⁡(−1−136)≈−0.875075474 rad2x = \arcsin\Big(\frac{-1-\sqrt{13}}6\Big) \approx -0.875075474\text{ rad}

The other solution in the range [−π,π][-\pi, \pi] is:

2x=−π−(−0.875075474)≈−2.266517178 rad2x = -\pi - (-0.875075474) \approx -2.266517178\text{ rad}

Dividing by 22 to find xx:

x= −0.8750754742≈−0.438 radx= −2.2665171782≈−1.13 rad\begin{align*} x =&\, \frac{-0.875075474}{2} \approx -0.438\text{ rad} \\ x =&\, \frac{-2.266517178}{2} \approx -1.13\text{ rad} \end{align*}

Final Solutions

The values of xx (rounded to 3 significant figures) are:

x≈−1.13 rad,−0.438 rad,0.225 rad,1.35 radx \approx -1.13\text{ rad}, \quad -0.438\text{ rad}, \quad 0.225\text{ rad}, \quad 1.35\text{ rad}