CalcGospel 國際數學圖譜
返回

IAL 2025 Oct Q9

A Level / Edexcel / P3

IAL 2025 Oct Paper · Question 9

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(a) Express

6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ6\sin^2\theta\cot 2\theta+4\sin\theta\cos\theta

in terms of sin⁡2θ\sin 2\theta and cos⁡2θ\cos 2\theta only.

(3)

(b) Hence show that the equation

3cot⁡2θ−14=6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ3\cot 2\theta-14=6\sin^2\theta\cot 2\theta+4\sin\theta\cos\theta

can be written in the form

5sin⁡22θ+14sin⁡2θ−3=05\sin^2 2\theta+14\sin 2\theta-3=0
(3)

(c) Hence solve, for 0<x<90∘0<x<90^\circ, the equation

3cot⁡2x−14=6sin⁡2xcot⁡2x+4sin⁡xcos⁡x3\cot 2x-14=6\sin^2x\cot 2x+4\sin x\cos x

giving your answers to one decimal place.

(3)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(a) 将

6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ6\sin^2\theta\cot 2\theta+4\sin\theta\cos\theta

化为只含 sin⁡2θ\sin 2\theta 与 cos⁡2θ\cos 2\theta 的形式。

(b) 由此证明方程

3cot⁡2θ−14=6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ3\cot 2\theta-14=6\sin^2\theta\cot 2\theta+4\sin\theta\cos\theta

可写成

5sin⁡22θ+14sin⁡2θ−3=05\sin^2 2\theta+14\sin 2\theta-3=0

的形式。

(c) 由此在 0<x<90∘0<x<90^\circ 内解方程

3cot⁡2x−14=6sin⁡2xcot⁡2x+4sin⁡xcos⁡x3\cot 2x-14=6\sin^2x\cot 2x+4\sin x\cos x

答案精确到小数点后 1 位。

解答

(a)

Use

sin⁡2θ=1−cos⁡2θ2\sin^2\theta=\frac{1-\cos2\theta}{2}

and

4sin⁡θcos⁡θ=2sin⁡2θ4\sin\theta\cos\theta=2\sin2\theta

Then

6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ=6(1−cos⁡2θ2)cot⁡2θ+2sin⁡2θ=3(1−cos⁡2θ)cot⁡2θ+2sin⁡2θ\begin{aligned} 6\sin^2\theta\cot2\theta+4\sin\theta\cos\theta &=6\left(\frac{1-\cos2\theta}{2}\right)\cot2\theta+2\sin2\theta \\ &=3(1-\cos2\theta)\cot2\theta+2\sin2\theta \end{aligned}

Since

cot⁡2θ=cos⁡2θsin⁡2θ\cot2\theta=\frac{\cos2\theta}{\sin2\theta}

we get

6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ=3(1−cos⁡2θ)cos⁡2θsin⁡2θ+2sin⁡2θ\boxed{ 6\sin^2\theta\cot2\theta+4\sin\theta\cos\theta =\frac{3(1-\cos2\theta)\cos2\theta}{\sin2\theta} +2\sin2\theta }

Equivalently,

6sin⁡2θcot⁡2θ+4sin⁡θcos⁡θ=(3−3cos⁡2θ)cos⁡2θsin⁡2θ+2sin⁡2θ\boxed{ 6\sin^2\theta\cot2\theta+4\sin\theta\cos\theta =\frac{(3-3\cos2\theta)\cos2\theta}{\sin2\theta} +2\sin2\theta }

This is in terms of sin⁡2θ\sin2\theta and cos⁡2θ\cos2\theta only.

(b)

From part (a), the equation becomes

3cot⁡2θ−14=(3−3cos⁡2θ)cos⁡2θsin⁡2θ+2sin⁡2θ3\cot2\theta-14 =\frac{(3-3\cos2\theta)\cos2\theta}{\sin2\theta} +2\sin2\theta

Write the left side using cot⁡2θ=cos⁡2θsin⁡2θ\cot2\theta=\frac{\cos2\theta}{\sin2\theta}:

3cos⁡2θsin⁡2θ−14=(3−3cos⁡2θ)cos⁡2θsin⁡2θ+2sin⁡2θ\frac{3\cos2\theta}{\sin2\theta}-14 =\frac{(3-3\cos2\theta)\cos2\theta}{\sin2\theta} +2\sin2\theta

Multiply by sin⁡2θ\sin2\theta:

3cos⁡2θ−14sin⁡2θ=(3−3cos⁡2θ)cos⁡2θ+2sin⁡22θ3\cos2\theta-14\sin2\theta =(3-3\cos2\theta)\cos2\theta+2\sin^2 2\theta

Expand the right side:

3cos⁡2θ−14sin⁡2θ=3cos⁡2θ−3cos⁡22θ+2sin⁡22θ3\cos2\theta-14\sin2\theta =3\cos2\theta-3\cos^2 2\theta+2\sin^2 2\theta

Cancel 3cos⁡2θ3\cos2\theta from both sides:

−14sin⁡2θ=−3cos⁡22θ+2sin⁡22θ-14\sin2\theta=-3\cos^2 2\theta+2\sin^2 2\theta

Use

cos⁡22θ=1−sin⁡22θ\cos^2 2\theta=1-\sin^2 2\theta

Then

−14sin⁡2θ=−3(1−sin⁡22θ)+2sin⁡22θ=−3+5sin⁡22θ\begin{aligned} -14\sin2\theta &=-3(1-\sin^2 2\theta)+2\sin^2 2\theta \\ &=-3+5\sin^2 2\theta \end{aligned}

Hence

5sin⁡22θ+14sin⁡2θ−3=05\sin^2 2\theta+14\sin2\theta-3=0

as required.

(c)

Using part (b), solve

5sin⁡22x+14sin⁡2x−3=05\sin^2 2x+14\sin2x-3=0

Let

u=sin⁡2xu=\sin2x

Then

5u2+14u−3=05u^2+14u-3=0

Factorise:

5u2+14u−3=(5u−1)(u+3)5u^2+14u-3=(5u-1)(u+3)

So

u=15oru=−3u=\frac15\quad\text{or}\quad u=-3

Since u=sin⁡2xu=\sin2x, we must have −1≤u≤1-1\le u\le 1, so u=−3u=-3 is impossible.

Therefore

sin⁡2x=15\sin2x=\frac15

Given

0<x<90∘0<x<90^\circ

we have

0<2x<180∘0<2x<180^\circ

In this interval, sin⁡2x=15\sin2x=\frac15 has two solutions:

2x=sin⁡−1(15)2x=\sin^{-1}\left(\frac15\right)

or

2x=180∘−sin⁡−1(15)2x=180^\circ-\sin^{-1}\left(\frac15\right)

So

x=5.768…∘x=5.768\ldots^\circ

or

x=84.231…∘x=84.231\ldots^\circ

Therefore, to one decimal place,

x=5.8∘, 84.2∘\boxed{x=5.8^\circ,\ 84.2^\circ}