题目

Problem

In this question you should show detailed reasoning.

Solutions relying entirely on calculator technology are not acceptable.

(a) Show that the equation

2\sin(\theta-30^\circ)=5\cos\theta

can be written in the form

\tan\theta=2\sqrt{3}

(4)

(b) Hence, or otherwise, solve for $0\le x\le 360^\circ$

2\sin(x-10^\circ)=5\cos(x+20^\circ)

giving your answers to one decimal place.

(3)

题目中文翻译

本题中你应写出详细的推理过程。

不接受完全依赖计算器技术的解法。

(a) 证明方程

2\sin(\theta-30^\circ)=5\cos\theta

可以写成

\tan\theta=2\sqrt{3}

的形式。

(b) 由此，或用其他方法，在 $0\le x\le 360^\circ$ 内解方程

2\sin(x-10^\circ)=5\cos(x+20^\circ)

答案精确到小数点后 1 位。

解答

Start with

2\sin(\theta-30^\circ)=5\cos\theta

Using

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

we get

2(\sin\theta\cos30^\circ-\cos\theta\sin30^\circ)=5\cos\theta

Since

\cos30^\circ=\frac{\sqrt3}{2}, \qquad \sin30^\circ=\frac12

we have

2\left(\frac{\sqrt3}{2}\sin\theta-\frac12\cos\theta\right)=5\cos\theta

\sqrt3\sin\theta-\cos\theta=5\cos\theta

Hence

\sqrt3\sin\theta=6\cos\theta

Dividing by $\cos\theta$ ,

\sqrt3\tan\theta=6

Therefore

\tan\theta=\frac{6}{\sqrt3}=2\sqrt3

as required.

The equation is

2\sin(x-10^\circ)=5\cos(x+20^\circ)

Let

\theta=x+20^\circ

Then

x-10^\circ=\theta-30^\circ

So the equation becomes

2\sin(\theta-30^\circ)=5\cos\theta

From part (a),

\tan\theta=2\sqrt3

Thus

\theta=73.897\ldots^\circ,\ 253.897\ldots^\circ

for $0\le x\le 360^\circ$ .

Since

x=\theta-20^\circ

we get

x=53.897\ldots^\circ,\ 233.897\ldots^\circ

Therefore, to one decimal place,

\boxed{x=53.9^\circ,\ 233.9^\circ}