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P2 2023 Jan Q8

A Level / Edexcel / P2

IAL 2023 Jan Paper · Question 8

题目

Problem

In this question you must show all stages of your working.

Solutions based entirely on calculator technology are not acceptable.

(i) Solve, for −π2<x<π-\frac{\pi}{2} < x < \pi, the equation

5sin⁡(3x+0.1)+2=05 \sin(3x + 0.1) + 2 = 0

giving your answers, in radians, to 2 decimal places.

(4)

(ii) Solve, for 0<θ<360∘0 < \theta < 360^{\circ}, the equation

2tan⁡θsin⁡θ=5+cos⁡θ2 \tan\theta \sin\theta = 5 + \cos\theta

giving your answers, in degrees, to one decimal place.

(5)
题目中文翻译

本题必须展示全部计算步骤。完全依赖计算器技术的解答不予接受。

(i) 在 −π/2<x<π-\pi/2<x<\pi 范围内解方程

5sin⁡(3x+0.1)+2=0,5\sin(3x+0.1)+2=0,

答案用弧度表示，保留两位小数。（4 分）

(ii) 在 0<θ<360∘0<\theta<360^\circ 范围内解方程

2tan⁡θsin⁡θ=5+cos⁡θ,2\tan\theta\sin\theta=5+\cos\theta,

答案用度表示，保留一位小数。（5 分）

解答

(i)

解法一

思路

展开

先化简为 sin⁡(3x+0.1)=−0.4\sin(3x+0.1)=-0.4。设 y=3x+0.1y=3x+0.1，题目范围 −π/2<x<π-\pi/2<x<\pi 对应

−3π2+0.1<y<3π+0.1.-\frac{3\pi}{2}+0.1<y<3\pi+0.1.

设主值 α=arcsin⁡(−0.4)\alpha=\arcsin(-0.4)。所有角来自 y=α+2kπy=\alpha+2k\pi 或 y=π−α+2kπy=\pi-\alpha+2k\pi，其中 kk 为整数。只保留上述区间内的四个角，再代入 x=(y−0.1)/3x=(y-0.1)/3。计算时保留未舍入的 α\alpha，最后才把 xx 舍入到两位小数。

答题过程

展开 5sin⁡(3x+0.1)+2= 0sin⁡(3x+0.1)= −0.4.\begin{align*} 5\sin(3x+0.1)+2=&\,0\\ \sin(3x+0.1)=&\,-0.4. \end{align*}

Let y=3x+0.1y=3x+0.1. Since −π/2<x<π-\pi/2<x<\pi,

−3π2+0.1<y<3π+0.1,-\frac{3\pi}{2}+0.1<y<3\pi+0.1,

or approximately −4.61239<y<9.52478-4.61239<y<9.52478.

Let α=arcsin⁡(−0.4)=−0.411516846…\alpha=\arcsin(-0.4)=-0.411516846\ldots. In this interval, the solutions are

y= −π−α,α, π−α,2π+α.\begin{align*} y=&\,-\pi-\alpha,\quad \alpha,\\ &\,\pi-\alpha,\quad 2\pi+\alpha. \end{align*}

Their approximate values are

y≈ −2.73008,−0.41152, 3.55311,5.87167.\begin{align*} y\approx&\,-2.73008,\quad -0.41152,\\ &\,3.55311,\quad 5.87167. \end{align*}

Using the unrounded angles in x=(y−0.1)/3x=(y-0.1)/3 gives

x≈ −0.943358603,−0.170505615, 1.151036500,1.923889487.\begin{align*} x\approx&\,-0.943358603,\quad -0.170505615,\\ &\,1.151036500,\quad 1.923889487. \end{align*}

Therefore, to 2 decimal places, the solutions in radians are

x=−0.94,−0.17,1.15,1.92.x=-0.94,\quad -0.17,\quad 1.15,\quad 1.92.

(ii)

解法一

思路

展开

1. 改写正切函数：利用关系式 tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}。

2(sin⁡θcos⁡θ)sin⁡θ= 5+cos⁡θ2sin⁡2θcos⁡θ= 5+cos⁡θ.\begin{align*} 2 \Big( \frac{\sin\theta}{\cos\theta} \Big) \sin\theta =&\, 5 + \cos\theta \\ \frac{2\sin^2\theta}{\cos\theta} =&\, 5 + \cos\theta. \end{align*}

2. 去分母与统一三角函数： 方程两边同乘以 cos⁡θ\cos\theta（需注意 cos⁡θ≠0\cos\theta \neq 0）：

2sin⁡2θ=5cos⁡θ+cos⁡2θ2 \sin^2\theta = 5 \cos\theta + \cos^2\theta

利用 sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta 将正弦转换为余弦：

2(1−cos⁡2θ)= 5cos⁡θ+cos⁡2θ3cos⁡2θ+5cos⁡θ−2= 0.\begin{align*} 2(1 - \cos^2\theta) =&\, 5 \cos\theta + \cos^2\theta \\ 3\cos^2\theta + 5\cos\theta - 2 =&\, 0. \end{align*}

3. 因式分解求解： 将二次三项式因式分解为 (3cos⁡θ−1)(cos⁡θ+2)=0(3\cos\theta - 1)(\cos\theta + 2) = 0。 由于余弦值的范围为 [−1,1][-1, 1]，所以 cos⁡θ=−2\cos\theta = -2 无解。 故 cos⁡θ=13\cos\theta = \frac{1}{3}。

4. 求 θ\theta 的解： 在范围 0<θ<360∘0 < \theta < 360^{\circ} 内，余弦值为正，有两个解：第一象限角和第四象限角。

答题过程

展开

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

2tan⁡θsin⁡θ= 5+cos⁡θ2(sin⁡θcos⁡θ)sin⁡θ= 5+cos⁡θ2sin⁡2θcos⁡θ= 5+cos⁡θ.\begin{align*} 2 \tan\theta \sin\theta =&\, 5 + \cos\theta \\ 2 \Big(\frac{\sin\theta}{\cos\theta}\Big) \sin\theta =&\, 5 + \cos\theta \\ \frac{2\sin^2\theta}{\cos\theta} =&\, 5 + \cos\theta. \end{align*}

Multiply by cos⁡θ\cos\theta:

2sin⁡2θ= 5cos⁡θ+cos⁡2θ.\begin{align*} 2\sin^2\theta =&\, 5\cos\theta + \cos^2\theta. \end{align*}

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

2(1−cos⁡2θ)= 5cos⁡θ+cos⁡2θ2−2cos⁡2θ= 5cos⁡θ+cos⁡2θ3cos⁡2θ+5cos⁡θ−2= 0(3cos⁡θ−1)(cos⁡θ+2)= 0.\begin{align*} 2(1 - \cos^2\theta) =&\, 5\cos\theta + \cos^2\theta \\ 2 - 2\cos^2\theta =&\, 5\cos\theta + \cos^2\theta \\ 3\cos^2\theta + 5\cos\theta - 2 =&\, 0 \\ (3\cos\theta - 1)(\cos\theta + 2) =&\, 0. \end{align*}

Since −1≤cos⁡θ≤1-1 \le \cos\theta \le 1, cos⁡θ=−2\cos\theta = -2 has no real solutions.

Thus, we have:

3cos⁡θ−1= 0cos⁡θ= 13.\begin{align*} 3\cos\theta - 1 =&\, 0 \\ \cos\theta =&\, \frac{1}{3}. \end{align*}

For 0<θ<360∘0 < \theta < 360^{\circ}:

θ= arccos⁡(13)≈ 70.5288∘≈ 70.5∘,\begin{align*} \theta =&\, \arccos\Big(\frac{1}{3}\Big) \\ \approx&\, 70.5288^{\circ} \\ \approx&\, 70.5^{\circ}, \end{align*}

and the other solution is:

θ= 360∘−70.5288∘≈ 289.4712∘≈ 289.5∘.\begin{align*} \theta =&\, 360^{\circ} - 70.5288^{\circ} \\ \approx&\, 289.4712^{\circ} \\ \approx&\, 289.5^{\circ}. \end{align*}

Therefore, the solutions are:

θ= 70.5∘, 289.5∘(to 1 d.p.).\begin{align*} \theta =&\, 70.5^{\circ}, \, 289.5^{\circ} \quad \text{(to 1 d.p.)}. \end{align*}