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IAL 2022 May Q9

A Level / Edexcel / P1

IAL 2022 May Paper · Question 9

题目

Problem

Figure 4 shows part of the graph of the curve with equation y=sin⁡xy=\sin x.

Figure 4

Given that sin⁡α=p\sin\alpha=p, where 0<α<90∘0<\alpha<90^\circ.

(a) state, in terms of pp, the value of

(i) 2sin⁡(180∘−α)2\sin(180^\circ-\alpha)

(ii) sin⁡(α−180∘)\sin(\alpha-180^\circ)

(iii) 3+sin⁡(180∘+α)3+\sin(180^\circ+\alpha)

(3)

A copy of Figure 4, labelled Diagram 1, is shown on page 27.

Diagram 1

On Diagram 1,

(b) sketch the graph of y=sin⁡2xy=\sin2x.

(2)

(c) Hence find, in terms of α\alpha, the xx coordinates of any points in the interval 0<x<180∘0<x<180^\circ where

sin⁡2x=p.\begin{align*} \sin2x=p. \end{align*}
(3)

解答

(a)

解法一

思路

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利用正弦图像的对称性：sin⁡(180∘−α)=sin⁡α\sin(180^\circ-\alpha)=\sin\alpha，而 sin⁡(180∘+α)=−sin⁡α\sin(180^\circ+\alpha)=-\sin\alpha。

答题过程

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Since sin⁡α=p\sin\alpha=p,

2sin⁡(180∘−α)= 2sin⁡α= 2p.\begin{align*} 2\sin(180^\circ-\alpha) =&\,2\sin\alpha\\ =&\,2p. \end{align*}

Also,

sin⁡(α−180∘)=−p.\begin{align*} \sin(\alpha-180^\circ)=-p. \end{align*}

Finally,

3+sin⁡(180∘+α)= 3−sin⁡α= 3−p.\begin{align*} 3+\sin(180^\circ+\alpha) =&\,3-\sin\alpha\\ =&\,3-p. \end{align*}

(b)

解法一

思路

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y=sin⁡2xy=\sin2x 的周期是 180∘180^\circ，振幅仍是 11。因此在同样的横向范围内，它比 y=sin⁡xy=\sin x 完成更多次循环。

答题过程

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The graph of y=sin⁡2xy=\sin2x has the same amplitude as y=sin⁡xy=\sin x, but its period is

180∘.\begin{align*} 180^\circ. \end{align*}

It passes through the xx-axis at

x=0∘, 90∘, 180∘, 270∘,…\begin{align*} x=0^\circ,\ 90^\circ,\ 180^\circ,\ 270^\circ,\ldots \end{align*}

with the corresponding sine-wave shape.

(c)

解法一

思路

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令 u=2xu=2x，则 sin⁡u=p=sin⁡α\sin u=p=\sin\alpha。在 0<x<180∘0<x<180^\circ 中，0<u<360∘0<u<360^\circ，所以 u=αu=\alpha 或 u=180∘−αu=180^\circ-\alpha。

答题过程

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Let

u=2x.\begin{align*} u=2x. \end{align*}

Since 0<x<180∘0<x<180^\circ,

0<u<360∘.\begin{align*} 0<u<360^\circ. \end{align*}

The solutions of

sin⁡u=p=sin⁡α\begin{align*} \sin u=p=\sin\alpha \end{align*}

in this interval are

u=αoru=180∘−α.\begin{align*} u=\alpha \quad\text{or}\quad u=180^\circ-\alpha. \end{align*}

Therefore

2x= αor2x=180∘−α.\begin{align*} 2x=&\,\alpha \quad\text{or}\quad 2x=180^\circ-\alpha. \end{align*}

So

x=α2orx=90∘−α2.\begin{align*} x=\frac{\alpha}{2} \quad\text{or}\quad x=90^\circ-\frac{\alpha}{2}. \end{align*}