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IAL 2025 May Q8

A Level / Edexcel / P3

IAL 2025 May Paper · Question 8

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(a) Prove that

tan⁡3x≡3tan⁡x−tan⁡3x1−3tan⁡2xx≠(2n+1)π6, n∈Z\tan 3x\equiv \frac{3\tan x-\tan^3x}{1-3\tan^2x}\qquad x\ne (2n+1)\frac{\pi}{6},\ n\in\mathbb{Z}
(3)

(b) Hence solve, for 0<θ<π20<\theta<\dfrac{\pi}{2},

3tan⁡θ−tan⁡3θ1−3tan⁡2θ=2sec⁡23θ−8\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=2\sec^2 3\theta-8

giving your answers to 2 decimal places.

(5)
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) 证明

tan⁡3x≡3tan⁡x−tan⁡3x1−3tan⁡2xx≠(2n+1)π6, n∈Z\tan 3x\equiv \frac{3\tan x-\tan^3x}{1-3\tan^2x}\qquad x\ne (2n+1)\frac{\pi}{6},\ n\in\mathbb{Z}

(b) 由此在 0<θ<π20<\theta<\dfrac{\pi}{2} 内解方程

3tan⁡θ−tan⁡3θ1−3tan⁡2θ=2sec⁡23θ−8\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=2\sec^2 3\theta-8

答案保留到小数点后 2 位。

解答

(a)

Use the addition formula

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}

Write

tan⁡3x=tan⁡(2x+x)\tan3x=\tan(2x+x)

Then

tan⁡3x=tan⁡2x+tan⁡x1−tan⁡2xtan⁡x\tan3x =\frac{\tan2x+\tan x}{1-\tan2x\tan x}

Also,

tan⁡2x=2tan⁡x1−tan⁡2x\tan2x=\frac{2\tan x}{1-\tan^2x}

Substitute this into the expression for tan⁡3x\tan3x:

tan⁡3x=2tan⁡x1−tan⁡2x+tan⁡x1−2tan⁡x1−tan⁡2xtan⁡x\tan3x =\frac{\frac{2\tan x}{1-\tan^2x}+\tan x} {1-\frac{2\tan x}{1-\tan^2x}\tan x}

Simplify the numerator:

2tan⁡x1−tan⁡2x+tan⁡x=2tan⁡x+tan⁡x(1−tan⁡2x)1−tan⁡2x=3tan⁡x−tan⁡3x1−tan⁡2x\frac{2\tan x}{1-\tan^2x}+\tan x =\frac{2\tan x+\tan x(1-\tan^2x)}{1-\tan^2x} =\frac{3\tan x-\tan^3x}{1-\tan^2x}

Simplify the denominator:

1−2tan⁡2x1−tan⁡2x=1−tan⁡2x−2tan⁡2x1−tan⁡2x=1−3tan⁡2x1−tan⁡2x1-\frac{2\tan^2x}{1-\tan^2x} =\frac{1-\tan^2x-2\tan^2x}{1-\tan^2x} =\frac{1-3\tan^2x}{1-\tan^2x}

Therefore

tan⁡3x=3tan⁡x−tan⁡3x1−tan⁡2x1−3tan⁡2x1−tan⁡2x=3tan⁡x−tan⁡3x1−3tan⁡2x\begin{aligned} \tan3x &=\frac{\frac{3\tan x-\tan^3x}{1-\tan^2x}} {\frac{1-3\tan^2x}{1-\tan^2x}} \\ &=\frac{3\tan x-\tan^3x}{1-3\tan^2x} \end{aligned}

Hence

tan⁡3x≡3tan⁡x−tan⁡3x1−3tan⁡2x\boxed{\tan 3x\equiv \frac{3\tan x-\tan^3x}{1-3\tan^2x}}

(b)

Using part (a),

3tan⁡θ−tan⁡3θ1−3tan⁡2θ=tan⁡3θ\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} =\tan3\theta

So the equation becomes

tan⁡3θ=2sec⁡23θ−8\tan3\theta=2\sec^2 3\theta-8

Use

sec⁡23θ=1+tan⁡23θ\sec^2 3\theta=1+\tan^2 3\theta

Then

tan⁡3θ=2(1+tan⁡23θ)−8\tan3\theta=2(1+\tan^2 3\theta)-8

Rearrange:

2tan⁡23θ−tan⁡3θ−6=02\tan^2 3\theta-\tan3\theta-6=0

Factorise:

(2tan⁡3θ+3)(tan⁡3θ−2)=0(2\tan3\theta+3)(\tan3\theta-2)=0

So

tan⁡3θ=−32\tan3\theta=-\frac32

or

tan⁡3θ=2\tan3\theta=2

Given

0<θ<π20<\theta<\frac{\pi}{2}

we have

0<3θ<3π20<3\theta<\frac{3\pi}{2}

For tan⁡3θ=2\tan3\theta=2,

3θ=tan⁡−123\theta=\tan^{-1}2

or

3θ=π+tan⁡−123\theta=\pi+\tan^{-1}2

For tan⁡3θ=−32\tan3\theta=-\frac32, the solution in this interval is

3θ=π−tan⁡−1323\theta=\pi-\tan^{-1}\frac32

Therefore

θ=0.3688…,0.7196…,1.4160…\theta=0.3688\ldots,\quad 0.7196\ldots,\quad 1.4160\ldots

So, to 2 decimal places,

θ=0.37, 0.72, 1.42\boxed{\theta=0.37,\ 0.72,\ 1.42}