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IAL 2025 May A Q6

A Level / Edexcel / P2

IAL 2025 May A Paper · Question 6

题目

Problem

(i) Given that θ\theta is measured in degrees and

  • cos⁡θ=15\cos\theta=\frac15
  • 180∘<θ<360∘180^\circ<\theta<360^\circ

use trigonometric identities to find the exact value of

(a) sin⁡θ\sin\theta

(b) tan⁡θ\tan\theta

giving the answers as fully simplified surds where appropriate.

(4)

(ii)

Figure 2

The height of sea water, hh metres, on a harbour wall, tt hours after midnight on a particular day is given by

h=4+3cos⁡(30t−40)∘0≤t<24h=4+3\cos(30t-40)^\circ \qquad 0\leq t<24

A sketch of hh against tt is shown in Figure 2.

(a) Find the minimum height of sea water on the harbour wall.

(1)

(b) Find the exact time of day when this minimum height first occurs.

(3)

When t=Tt=T, as shown in Figure 2, a boat enters the harbour when the height of sea water on the harbour wall is 3.53.5 m.

(c) Use Figure 2 and the given equation to find the value of TT to 22 decimal places.

(4)

解答

(i)(a)

解法一

思路

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用 sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1。因为 θ\theta 在第三或第四象限，而 cos⁡θ>0\cos\theta>0，所以 θ\theta 在第四象限，sin⁡θ<0\sin\theta<0。

答题过程

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Using

sin⁡2θ+cos⁡2θ=1,\begin{align*} \sin^2\theta+\cos^2\theta=1, \end{align*}

and cos⁡θ=15\cos\theta=\frac15,

sin⁡2θ+(15)2= 1sin⁡2θ= 1−125= 2425.\begin{align*} \sin^2\theta+\left(\frac15\right)^2=&\,1\\ \sin^2\theta=&\,1-\frac1{25}\\ =&\,\frac{24}{25}. \end{align*}

Since 180∘<θ<360∘180^\circ<\theta<360^\circ and cos⁡θ>0\cos\theta>0, θ\theta is in the fourth quadrant. Hence sin⁡θ<0\sin\theta<0.

Therefore,

sin⁡θ=−245=−265.\begin{align*} \sin\theta=-\frac{\sqrt{24}}5=-\frac{2\sqrt6}{5}. \end{align*}

解法二

思路

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把 cos⁡θ=15\cos\theta=\frac15 看成参考直角三角形中的 adjacent 和 hypotenuse。先用 Pythagoras’ theorem 找 opposite，再根据象限决定 sin⁡θ\sin\theta 的正负。

答题过程

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For the reference triangle, take the adjacent side as 11 and the hypotenuse as 55.

By Pythagoras’ theorem, the opposite side has length

52−12=24=26.\begin{align*} \sqrt{5^2-1^2}=\sqrt{24}=2\sqrt6. \end{align*}

Since 180∘<θ<360∘180^\circ<\theta<360^\circ and cos⁡θ>0\cos\theta>0, θ\theta is in the fourth quadrant.

Therefore sin⁡θ<0\sin\theta<0, so

sin⁡θ=−265.\begin{align*} \sin\theta=-\frac{2\sqrt6}{5}. \end{align*}

(i)(b)

解法一

思路

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用 tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}，直接代入上一小题结果。

答题过程

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Using

tan⁡θ=sin⁡θcos⁡θ,\begin{align*} \tan\theta=\frac{\sin\theta}{\cos\theta}, \end{align*}

we get

tan⁡θ= −26515= −26.\begin{align*} \tan\theta =&\,\frac{-\frac{2\sqrt6}{5}}{\frac15}\\ =&\,-2\sqrt6. \end{align*}

解法二

思路

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沿用参考三角形。第四象限中 tan⁡θ\tan\theta 为负，而大小是 oppositeadjacent\frac{\text{opposite}}{\text{adjacent}}。

答题过程

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From the reference triangle, the opposite side has length 262\sqrt6 and the adjacent side has length 11.

Since θ\theta is in the fourth quadrant, tan⁡θ<0\tan\theta<0.

Therefore,

tan⁡θ=−261=−26.\begin{align*} \tan\theta=-\frac{2\sqrt6}{1}=-2\sqrt6. \end{align*}

(ii)(a)

解法一

思路

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cos⁡\cos 的最小值是 −1-1，所以 hh 的最小值是 4−34-3。

答题过程

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The minimum value of cos⁡(30t−40)∘\cos(30t-40)^\circ is −1-1.

So the minimum height is

4+3(−1)=1.\begin{align*} 4+3(-1)=1. \end{align*}

Therefore, the minimum height is

1 m.\begin{align*} 1\text{ m}. \end{align*}

(ii)(b)

解法一

思路

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最小值第一次出现时，cos⁡(30t−40)∘=−1\cos(30t-40)^\circ=-1，也就是角度第一次等于 180∘180^\circ。

答题过程

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The first minimum occurs when

30t−40=180.\begin{align*} 30t-40=180. \end{align*}

So

30t= 220t= 223= 7+13.\begin{align*} 30t=&\,220\\ t=&\,\frac{22}{3}\\ =&\,7+\frac13. \end{align*}

This is 77 hours and 2020 minutes after midnight.

Therefore, the exact time is

7:20 am.\begin{align*} 7{:}20\text{ am}. \end{align*}

(ii)(c)

解法一

思路

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令 h=3.5h=3.5。从图中位置看，TT 是最低点之后、曲线上升后对应的那个解，所以要选较晚的角度。

答题过程

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When h=3.5h=3.5,

4+3cos⁡(30T−40)∘= 3.53cos⁡(30T−40)∘= −0.5cos⁡(30T−40)∘= −16.\begin{align*} 4+3\cos(30T-40)^\circ=&\,3.5\\ 3\cos(30T-40)^\circ=&\,-0.5\\ \cos(30T-40)^\circ=&\,-\frac16. \end{align*}

The relevant solution from the graph is

30T−40=459.6…∘.\begin{align*} 30T-40=459.6\ldots^\circ. \end{align*}

Thus

30T= 499.6…T= 16.653….\begin{align*} 30T=&\,499.6\ldots\\ T=&\,16.653\ldots. \end{align*}

Therefore,

T=16.65.\begin{align*} T=16.65. \end{align*}