题目

Problem

Figure 4 shows part of the curve with equation

\begin{align*} y=A\cos(x-30)^\circ \end{align*}

Figure 4

where $A$ is a constant.

The point $P$ is a minimum point on the curve and has coordinates $(30,-3)$ as shown in Figure 4.

(a) Write down the value of $A$ .

(1)

The point $Q$ is shown in Figure 4 and is a maximum point.

(b) Find the coordinates of $Q$ .

(3)

解答

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当 $x=30$ 时， $\cos(x-30)^\circ=\cos0^\circ=1$ ，所以此时 $y=A$ 。题目给出这个点是 $(30,-3)$ 。

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At $P(30,-3)$ ,

\begin{align*} y=&\,A\cos(30-30)^\circ\\ =&\,A\cos0^\circ\\ =&\,A. \end{align*}

Since $y=-3$ at $P$ ,

\begin{align*} A=-3. \end{align*}

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因为 $A=-3$ ，图像是普通 cosine 图像上下翻转并放大到振幅 $3$ 。最大值为 $3$ 。最大点发生在 $\cos(x-30)^\circ=-1$ 的位置。

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From part (a),

\begin{align*} y=-3\cos(x-30)^\circ. \end{align*}

The maximum value is

\begin{align*} y=3. \end{align*}

This occurs when

\begin{align*} \cos(x-30)^\circ=-1. \end{align*}

\begin{align*} x-30=&\,180+360n\\ x=&\,210+360n. \end{align*}

For the maximum point $Q$ shown in the figure,

\begin{align*} x=210+2(360)=930. \end{align*}

Therefore

\begin{align*} Q=(930,3). \end{align*}