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IAL 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)

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⁡θ  ⟹  tan⁡θcos⁡θ=sin⁡θcos⁡θcos⁡θ=sin⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta} \implies \tan\theta\cos\theta = \frac{\sin\theta}{\cos\theta}\cos\theta = \sin\theta

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=2  ⟹  3sin⁡2θ+sin⁡θ−1=03 - 3\sin^2\theta - \sin\theta = 2 \implies 3\sin^2\theta + \sin\theta - 1 = 0

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 - \left(\frac{\sin\theta}{\cos\theta}\right)\cos\theta =&\,\, 2 \\[2mm] 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 \\[2mm] 3 - 3\sin^2\theta - \sin\theta =&\,\, 2 \\[2mm] -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

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根据 (a) 部分的代数化简结论，对于方程：

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

我们可以直接将自变量 θ\theta 替换为 2x2x。因此，此方程可以等价写为关于 sin⁡2x\sin 2x 的二次方程：

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

设 y=sin⁡2xy = \sin 2x，则方程为 3y2+y−1=03y^2 + y - 1 = 0。

  1. 第一步：求解二次方程 由于无法直接因式分解，我们使用求根公式（quadratic formula）：

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

    计算其近似小数值：

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

  2. 第二步：确定求和区间的自变量范围 题目限制 xx 范围为 −π2⩽x⩽π2-\dfrac{\pi}{2} \leqslant x \leqslant \dfrac{\pi}{2}。 因为自变量是 2x2x，对应范围应乘以 2：

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

  3. 第三步：求解 2x2x 并求出 xx 我们需要在区间 [−π,π][-\pi, \pi] 中找出正弦值等于上述两个根的所有解，然后再除以 2 得到 xx 的取值。

    • 对于 sin⁡2x≈0.43426\sin 2x \approx 0.43426：

      • 主值（Principal Value）：2x1=arcsin⁡(0.43426)≈0.44924 rad2x_1 = \arcsin(0.43426) \approx 0.44924\text{ rad}。
      • 第二个解：2x2=π−0.44924≈2.69235 rad2x_2 = \pi - 0.44924 \approx 2.69235\text{ rad}。
      • 对应 xx 值为： x1≈0.225 rad,x2≈1.35 radx_1 \approx 0.225\text{ rad},\quad x_2 \approx 1.35\text{ rad}

    • 对于 sin⁡2x≈−0.76759\sin 2x \approx -0.76759：

      • 主值（Principal Value）：2x3=arcsin⁡(−0.76759)≈−0.87524 rad2x_3 = \arcsin(-0.76759) \approx -0.87524\text{ rad}。
      • 第二个解（区间在 [−π,π][-\pi, \pi]）：2x4=−π−(−0.87524)≈−2.26635 rad2x_4 = -\pi - (-0.87524) \approx -2.26635\text{ rad}。
      • 对应 xx 值为： x3≈−0.438 rad,x4≈−1.13 radx_3 \approx -0.438\text{ rad},\quad x_4 \approx -1.13\text{ rad}

结合以上，本题共有 4 个满足区间的实数解。

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)} \\[2mm] =&\,\, \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⁡(0.43426...)≈0.44924 rad2x = \arcsin(0.43426...) \approx 0.44924\text{ rad}

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

2x=π−0.44924≈2.69235 rad2x = \pi - 0.44924 \approx 2.69235\text{ rad}

Dividing by 22 to find xx:

x=  0.449242≈0.225 radx=  2.692352≈1.35 rad\begin{align*} x =&\,\, \frac{0.44924}{2} \approx 0.225\text{ rad} \\[2mm] x =&\,\, \frac{2.69235}{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⁡(−0.76759...)≈−0.87524 rad2x = \arcsin(-0.76759...) \approx -0.87524\text{ rad}

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

2x=−π−(−0.87524)≈−2.26635 rad2x = -\pi - (-0.87524) \approx -2.26635\text{ rad}

Dividing by 22 to find xx:

x=  −0.875242≈−0.438 radx=  −2.266352≈−1.13 rad\begin{align*} x =&\,\, \frac{-0.87524}{2} \approx -0.438\text{ rad} \\[2mm] x =&\,\, \frac{-2.26635}{2} \approx -1.13\text{ rad} \end{align*}

Final Solutions

The values of xx (rounded to 3 significant figures or 2 decimal places where appropriate) 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}