题目
Problem
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
(i) Given that θ \theta θ
cos θ = 1 5 \cos\theta=\frac1{\sqrt5} cos θ = 5 1 180 ∘ < θ < 360 ∘ 180^\circ<\theta<360^\circ 18 0 ∘ < θ < 36 0 ∘
use trigonometric identities to find the exact value of
(a) sin θ \sin\theta sin θ
(b) tan θ \tan\theta tan θ
giving the answers as fully simplified surds where appropriate.
(4)
(ii)
The height of sea water, h h h t t t
h = 4 + 3 cos ( 30 t − 40 ) ∘ 0 ≤ t < 24 \begin{align*}
h=&\,4+3\cos(30t-40)^\circ\\
&\,0\leq t<24
\end{align*} h = 4 + 3 cos ( 30 t − 40 ) ∘ 0 ≤ t < 24 A sketch of h h h t t t
(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 = T t=T t = T 3.5 3.5 3.5
(c) Use Figure 2 and the given equation to find the value of T T T 2 2 2
(4)
题目中文翻译
本题必须展示计算过程的所有步骤。
依赖计算器技术的解答不予接受。
(i) 已知 θ \theta θ
cos θ = 1 5 \cos\theta=\frac1{\sqrt5} cos θ = 5 1 180 ∘ < θ < 360 ∘ 180^\circ<\theta<360^\circ 18 0 ∘ < θ < 36 0 ∘
使用三角恒等式求以下各式的精确值：
(a) sin θ \sin\theta sin θ
(b) tan θ \tan\theta tan θ
适当时将答案写成完全化简的根式。
(4)
(ii) 在某一天，午夜后 t t t h h h
h = 4 + 3 cos ( 30 t − 40 ) ∘ h=4+3\cos(30t-40)^\circ h = 4 + 3 cos ( 30 t − 40 ) ∘ 其中 0 ≤ t < 24 0\leq t<24 0 ≤ t < 24
图2是 h h h t t t
(a) 求堤墙上海水的最小高度。
(1)
(b) 求一天中首次出现这个最小高度的准确时刻。
(3)
如图2所示，当 t = T t=T t = T 3.5 3.5 3.5
(c) 使用图2和给定方程求 T T T 2 2 2
(4)
解答
(i)(a)
解法一
思路
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用 sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1 θ \theta θ cos θ > 0 \cos\theta>0 cos θ > 0 θ \theta θ sin θ < 0 \sin\theta<0 sin θ < 0
答题过程
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Using
sin 2 θ + cos 2 θ = 1 , \begin{align*}
\sin^2\theta+\cos^2\theta=1,
\end{align*} sin 2 θ + cos 2 θ = 1 , and cos θ = 1 5 \cos\theta=\frac1{\sqrt5} cos θ = 5 1
sin 2 θ + ( 1 5 ) 2 = 1 sin 2 θ = 1 − 1 5 = 4 5 . \begin{align*}
\sin^2\theta+\bigg(\frac1{\sqrt5}\bigg)^2=&\,1\\
\sin^2\theta=&\,1-\frac15\\
=&\,\frac45.
\end{align*} sin 2 θ + ( 5 1 ) 2 = sin 2 θ = = 1 1 − 5 1 5 4 . Since 180 ∘ < θ < 360 ∘ 180^\circ<\theta<360^\circ 18 0 ∘ < θ < 36 0 ∘ cos θ > 0 \cos\theta>0 cos θ > 0 θ \theta θ sin θ < 0 \sin\theta<0 sin θ < 0
Therefore,
sin θ = − 2 5 = − 2 5 5 . \begin{align*}
\sin\theta=&\,-\frac2{\sqrt5}\\
=&\,-\frac{2\sqrt5}{5}.
\end{align*} sin θ = = − 5 2 − 5 2 5 .
解法二
思路
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把 cos θ = 1 5 \cos\theta=\frac1{\sqrt5} cos θ = 5 1 sin θ \sin\theta sin θ
答题过程
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For the reference triangle, take the adjacent side as 1 1 1 5 \sqrt5 5
By Pythagoras’ theorem, the opposite side has length
( 5 ) 2 − 1 2 = 5 − 1 = 2. \begin{align*}
\sqrt{(\sqrt5)^2-1^2}=&\,\sqrt{5-1}\\
=&\,2.
\end{align*} ( 5 ) 2 − 1 2 = = 5 − 1 2. Since 180 ∘ < θ < 360 ∘ 180^\circ<\theta<360^\circ 18 0 ∘ < θ < 36 0 ∘ cos θ > 0 \cos\theta>0 cos θ > 0 θ \theta θ
Therefore sin θ < 0 \sin\theta<0 sin θ < 0
sin θ = − 2 5 5 . \begin{align*}
\sin\theta=-\frac{2\sqrt5}{5}.
\end{align*} sin θ = − 5 2 5 .
(i)(b)
解法一
思路
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用 tan θ = sin θ cos θ \tan\theta=\frac{\sin\theta}{\cos\theta} tan θ = c o s θ s i n θ
答题过程
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Using
tan θ = sin θ cos θ , \begin{align*}
\tan\theta=\frac{\sin\theta}{\cos\theta},
\end{align*} tan θ = cos θ sin θ , we get
tan θ = − 2 5 5 ⋅ 5 = − 2 ⋅ 5 5 = − 2. \begin{align*}
\tan\theta
=&\,-\frac{2\sqrt5}{5}\cdot\sqrt5\\
=&\,-\frac{2\cdot5}{5}\\
=&\,-2.
\end{align*} tan θ = = = − 5 2 5 ⋅ 5 − 5 2 ⋅ 5 − 2.
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 tan θ opposite adjacent \frac{\text{opposite}}{\text{adjacent}} adjacent opposite
答题过程
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From the reference triangle, the opposite side has length 2 2 2 1 1 1
Since θ \theta θ tan θ < 0 \tan\theta<0 tan θ < 0
Therefore,
tan θ = − 2 1 = − 2. \begin{align*}
\tan\theta=-\frac21=-2.
\end{align*} tan θ = − 1 2 = − 2.
(ii)(a)
解法一
思路
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cos \cos cos − 1 -1 − 1 h h h 4 − 3 4-3 4 − 3
答题过程
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The minimum value of cos ( 30 t − 40 ) ∘ \cos(30t-40)^\circ cos ( 30 t − 40 ) ∘ − 1 -1 − 1
So the minimum height is
4 + 3 ( − 1 ) = 1. \begin{align*}
4+3(-1)=1.
\end{align*} 4 + 3 ( − 1 ) = 1. Therefore, the minimum height is
1 m . \begin{align*}
1\text{ m}.
\end{align*} 1 m .
(ii)(b)
解法一
思路
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最小值第一次出现时，cos ( 30 t − 40 ) ∘ = − 1 \cos(30t-40)^\circ=-1 cos ( 30 t − 40 ) ∘ = − 1 180 ∘ 180^\circ 18 0 ∘
答题过程
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For 0 ≤ t < 24 0\leq t<24 0 ≤ t < 24 u = 30 t − 40 u=30t-40 u = 30 t − 40 − 40 ≤ u < 680 -40\leq u<680 − 40 ≤ u < 680 u = 180 + 360 j u=180+360j u = 180 + 360 j j j j 180 ∘ 180^\circ 18 0 ∘
30 t − 40 = 180. \begin{align*}
30t-40=180.
\end{align*} 30 t − 40 = 180. So
30 t = 220 t = 22 3 = 7 + 1 3 . \begin{align*}
30t=&\,220\\
t=&\,\frac{22}{3}\\
=&\,7+\frac13.
\end{align*} 30 t = t = = 220 3 22 7 + 3 1 . This is 7 7 7 20 20 20
Therefore, the exact time is
7 : 20 am . \begin{align*}
7{:}20\text{ am}.
\end{align*} 7 : 20 am .
(ii)(c)
解法一
思路
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先令 h = 3.5 h=3.5 h = 3.5 T T T α + 360 ∘ \alpha+360^\circ α + 36 0 ∘
答题过程
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When h = 3.5 h=3.5 h = 3.5
4 + 3 cos ( 30 T − 40 ) ∘ = 3.5 3 cos ( 30 T − 40 ) ∘ = − 0.5 cos ( 30 T − 40 ) ∘ = − 1 6 . \begin{align*}
4+3\cos(30T-40)^\circ=&\,3.5\\
3\cos(30T-40)^\circ=&\,-0.5\\
\cos(30T-40)^\circ=&\,-\frac16.
\end{align*} 4 + 3 cos ( 30 T − 40 ) ∘ = 3 cos ( 30 T − 40 ) ∘ = cos ( 30 T − 40 ) ∘ = 3.5 − 0.5 − 6 1 . Let α \alpha α
cos α ∘ = − 1 6 . \cos\alpha^\circ=-\frac16. cos α ∘ = − 6 1 . Then
α = 99.594068 … . \alpha=99.594068\ldots. α = 99.594068 … . For 0 ≤ T < 24 0\leq T<24 0 ≤ T < 24 u = 30 T − 40 u=30T-40 u = 30 T − 40 − 40 ≤ u < 680 -40\leq u<680 − 40 ≤ u < 680
u 1 = α , u 2 = 360 − α , u 3 = 360 + α , u 4 = 720 − α . \begin{align*}
u_1=&\,\alpha,\\
u_2=&\,360-\alpha,\\
u_3=&\,360+\alpha,\\
u_4=&\,720-\alpha.
\end{align*} u 1 = u 2 = u 3 = u 4 = α , 360 − α , 360 + α , 720 − α . Figure 2 places T T T h = 3.5 h=3.5 h = 3.5
30 T − 40 = 360 + α T = 400 + α 30 = 16.653135 … . \begin{align*}
30T-40=&\,360+\alpha\\
T=&\,\frac{400+\alpha}{30}\\
=&\,16.653135\ldots.
\end{align*} 30 T − 40 = T = = 360 + α 30 400 + α 16.653135 … . Therefore,
T = 16.65 (to 2 decimal places) . T=16.65\quad\text{(to 2 decimal places)}. T = 16.65 (to 2 decimal places) . 
