题目

Problem

(a) Show that

\frac{1-\cos 2x}{2\sin 2x}\equiv k\tan x\qquad x\ne (90n)^\circ,\ n\in\mathbb{Z}

where $k$ is a constant to be found.

(3)

(b) Hence solve, for $0<\theta<90^\circ$ ,

\frac{9(1-\cos 2\theta)}{2\sin 2\theta}=2\sec^2\theta

giving your answers to one decimal place.

(Solutions based entirely on graphical or numerical methods are not acceptable.)

(6)

题目中文翻译

(a) 证明

\frac{1-\cos 2x}{2\sin 2x}\equiv k\tan x\qquad x\ne (90n)^\circ,\ n\in\mathbb{Z}

其中 $k$ 为待求常数。

(b) 由此在 $0<\theta<90^\circ$ 内解方程

\frac{9(1-\cos 2\theta)}{2\sin 2\theta}=2\sec^2\theta

答案精确到小数点后 1 位。

（不接受完全基于图像法或数值法的解答。）

解答

Use the identities

1-\cos2x=2\sin^2x

and

\sin2x=2\sin x\cos x

Then

\begin{aligned} \frac{1-\cos2x}{2\sin2x} &=\frac{2\sin^2x}{2(2\sin x\cos x)}\\ &=\frac{2\sin^2x}{4\sin x\cos x}\\ &=\frac12\cdot \frac{\sin x}{\cos x}\\ &=\frac12\tan x \end{aligned}

Therefore

\boxed{k=\frac12}

and

\boxed{\frac{1-\cos2x}{2\sin2x}\equiv \frac12\tan x}

Using part (a),

\frac{1-\cos2\theta}{2\sin2\theta}=\frac12\tan\theta

So the equation

\frac{9(1-\cos2\theta)}{2\sin2\theta}=2\sec^2\theta

becomes

9\left(\frac12\tan\theta\right)=2\sec^2\theta

Hence

\frac92\tan\theta=2\sec^2\theta

Multiply by $2$ :

9\tan\theta=4\sec^2\theta

Use

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

to get

9\tan\theta=4(1+\tan^2\theta)

4\tan^2\theta-9\tan\theta+4=0

Let

u=\tan\theta

Then

4u^2-9u+4=0

Using the quadratic formula:

u=\frac{9\pm\sqrt{(-9)^2-4(4)(4)}}{2(4)}

Thus

u=\frac{9\pm\sqrt{17}}{8}

\tan\theta=\frac{9+\sqrt{17}}{8}

\tan\theta=\frac{9-\sqrt{17}}{8}

Since

0<\theta<90^\circ

both values give valid solutions.

Therefore

\theta=58.633\ldots^\circ

\theta=31.367\ldots^\circ

Hence, to one decimal place,

\boxed{\theta=31.4^\circ,\ 58.6^\circ}