题目

Problem

In this question you must show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

(i) Solve, for $0<x\le \pi$ , the equation

2\sec^2x-3\tan x=2

giving the answers, as appropriate, to 3 significant figures.

(4)

(ii) Prove that

\frac{\sin 3\theta}{\sin\theta}-\frac{\cos 3\theta}{\cos\theta}\equiv 2

(4)

题目中文翻译

本题中你必须写出详细的推理过程。

不接受完全依赖计算器技术的解法。

(i) 在 $0<x\le \pi$ 内解方程

2\sec^2x-3\tan x=2

答案按需要保留 3 位有效数字。

(ii) 证明

\frac{\sin 3\theta}{\sin\theta}-\frac{\cos 3\theta}{\cos\theta}\equiv 2

解答

We need to solve

2\sec^2x-3\tan x=2

Using

\sec^2x=1+\tan^2x

we get

2(1+\tan^2x)-3\tan x=2

Expand:

2+2\tan^2x-3\tan x=2

2\tan^2x-3\tan x=0

Factorise:

\tan x(2\tan x-3)=0

Hence

\tan x=0

\tan x=\frac32

For

0<x\le\pi

the solution from $\tan x=0$ is

x=\pi

The solution from $\tan x=\dfrac32$ is

x=\arctan\frac32=0.9827\ldots

Therefore

\boxed{x=0.983,\ \pi}

Start with

\frac{\sin 3\theta}{\sin\theta} -\frac{\cos 3\theta}{\cos\theta}

Put the two fractions over a common denominator:

\frac{\sin 3\theta\cos\theta-\cos 3\theta\sin\theta} {\sin\theta\cos\theta}

Using

\sin A\cos B-\cos A\sin B=\sin(A-B)

the numerator becomes

\sin(3\theta-\theta)=\sin2\theta

Therefore

\frac{\sin 3\theta}{\sin\theta} -\frac{\cos 3\theta}{\cos\theta} \equiv \frac{\sin2\theta}{\sin\theta\cos\theta}

Now use

\sin2\theta=2\sin\theta\cos\theta

\frac{\sin2\theta}{\sin\theta\cos\theta} =\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} =2

Hence

\boxed{ \frac{\sin 3\theta}{\sin\theta} -\frac{\cos 3\theta}{\cos\theta} \equiv 2 }