题目
Problem
In this question you must show detailed reasoning.
Solutions relying entirely on calculator technology are not acceptable.
(i) Solve, for 0 ≤ x < 360 ∘ 0\leq x<360^\circ 0 ≤ x < 36 0 ∘
sin x tan x = 5 \sin x\tan x=5 sin x tan x = 5 giving your answers to one decimal place.
(6)
(ii)
Figure 1 shows a sketch of part of the curve with equation
y = A sin ( 2 θ − 3 π 8 ) + 2 y=A\sin\bigg(2\theta-\frac{3\pi}{8}\bigg)+2 y = A sin ( 2 θ − 8 3 π ) + 2 where A A A θ \theta θ
The points P P P Q Q Q R R R
Given that the y y y P P P 7 7 7
(a) state the value of A A A
(1)
(b) find the exact coordinates of Q Q Q
(3)
(c) find the value of θ \theta θ R R R 3 3 3
(4)
题目中文翻译
本题必须展示详细推理过程。不能完全依赖计算器技术。
(i) 解方程 sin x tan x = 5 \sin x\tan x=5 sin x tan x = 5 0 ≤ x < 360 ∘ 0\leq x<360^\circ 0 ≤ x < 36 0 ∘
（6分）
(ii) 图 1 显示曲线
y = A sin ( 2 θ − 3 π 8 ) + 2 y=A\sin\bigg(2\theta-\frac{3\pi}{8}\bigg)+2 y = A sin ( 2 θ − 8 3 π ) + 2 的一部分，其中 A A A θ \theta θ P P P Q Q Q R R R
已知 P P P y y y 7 7 7
(a) 写出 A A A
（1分）
(b) 求 Q Q Q
（3分）
(c) 求 R R R θ \theta θ
（4分）
解答
(i)
解法一
思路
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把 tan x \tan x tan x sin x cos x \frac{\sin x}{\cos x} c o s x s i n x sin 2 x = 1 − cos 2 x \sin^2x=1-\cos^2x sin 2 x = 1 − cos 2 x cos x \cos x cos x
答题过程
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Since
tan x = sin x cos x , \begin{align*}
\tan x=\frac{\sin x}{\cos x},
\end{align*} tan x = cos x sin x , the equation becomes
sin x ⋅ sin x cos x = 5 sin 2 x cos x = 5 sin 2 x = 5 cos x . \begin{align*}
\sin x\cdot\frac{\sin x}{\cos x}=&\,5\\
\frac{\sin^2x}{\cos x}=&\,5\\
\sin^2x=&\,5\cos x.
\end{align*} sin x ⋅ cos x sin x = cos x sin 2 x = sin 2 x = 5 5 5 cos x . Using sin 2 x = 1 − cos 2 x \sin^2x=1-\cos^2x sin 2 x = 1 − cos 2 x
1 − cos 2 x = 5 cos x cos 2 x + 5 cos x − 1 = 0. \begin{align*}
1-\cos^2x=&\,5\cos x\\
\cos^2x+5\cos x-1=&\,0.
\end{align*} 1 − cos 2 x = cos 2 x + 5 cos x − 1 = 5 cos x 0. Solve this quadratic in cos x \cos x cos x
cos x = − 5 ± 5 2 − 4 ( 1 ) ( − 1 ) 2 = − 5 ± 29 2 . \begin{align*}
\cos x=&\,\frac{-5\pm\sqrt{5^2-4(1)(-1)}}{2}\\
=&\,\frac{-5\pm\sqrt{29}}{2}.
\end{align*} cos x = = 2 − 5 ± 5 2 − 4 ( 1 ) ( − 1 ) 2 − 5 ± 29 . The value − 5 − 29 2 \frac{-5-\sqrt{29}}{2} 2 − 5 − 29 − 1 -1 − 1 cos x \cos x cos x
Thus
cos x = − 5 + 29 2 . \begin{align*}
\cos x=\frac{-5+\sqrt{29}}{2}.
\end{align*} cos x = 2 − 5 + 29 . For 0 ≤ x < 360 ∘ 0\leq x<360^\circ 0 ≤ x < 36 0 ∘
x = 78.9 ∘ , 281.1 ∘ . \begin{align*}
x=78.9^\circ,\ 281.1^\circ.
\end{align*} x = 78. 9 ∘ , 281. 1 ∘ .
同一二次式也可由 sin x tan x = tan 2 x cos x \sin x\tan x=\tan^2x\cos x sin x tan x = tan 2 x cos x tan 2 x = sec 2 x − 1 \tan^2x=\sec^2x-1 tan 2 x = sec 2 x − 1
( sec 2 x − 1 ) cos x = 5 1 cos x − cos x = 5 1 − cos 2 x = 5 cos x . \begin{align*}
(\sec^2x-1)\cos x =&\,5\\
\frac1{\cos x}-\cos x =&\,5\\
1-\cos^2x =&\,5\cos x.
\end{align*} ( sec 2 x − 1 ) cos x = cos x 1 − cos x = 1 − cos 2 x = 5 5 5 cos x . (ii)(a)
解法一
思路
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曲线的中线是 y = 2 y=2 y = 2 P P P y y y 7 7 7 7 − 2 = 5 7-2=5 7 − 2 = 5
答题过程
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The midline is
y = 2. \begin{align*}
y=2.
\end{align*} y = 2. Since the maximum value is 7 7 7
A = 7 − 2 = 5. \begin{align*}
A=7-2=5.
\end{align*} A = 7 − 2 = 5.
(ii)(b)
解法一
思路
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Q Q Q sin \sin sin − 1 -1 − 1 y y y 2 − 5 = − 3 2-5=-3 2 − 5 = − 3
答题过程
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At Q Q Q
sin ( 2 θ − 3 π 8 ) = − 1. \begin{align*}
\sin\bigg(2\theta-\frac{3\pi}{8}\bigg)=-1.
\end{align*} sin ( 2 θ − 8 3 π ) = − 1. So
2 θ − 3 π 8 = 3 π 2 . \begin{align*}
2\theta-\frac{3\pi}{8}=\frac{3\pi}{2}.
\end{align*} 2 θ − 8 3 π = 2 3 π . Hence
2 θ = 3 π 2 + 3 π 8 = 12 π 8 + 3 π 8 = 15 π 8 θ = 15 π 16 . \begin{align*}
2\theta=&\,\frac{3\pi}{2}+\frac{3\pi}{8}\\
=&\,\frac{12\pi}{8}+\frac{3\pi}{8}\\
=&\,\frac{15\pi}{8}\\
\theta=&\,\frac{15\pi}{16}.
\end{align*} 2 θ = = = θ = 2 3 π + 8 3 π 8 12 π + 8 3 π 8 15 π 16 15 π . The y y y
2 − 5 = − 3. \begin{align*}
2-5=-3.
\end{align*} 2 − 5 = − 3. Therefore,
Q ( 15 π 16 , − 3 ) . \begin{align*}
Q\bigg(\frac{15\pi}{16},-3\bigg).
\end{align*} Q ( 16 15 π , − 3 ) .
(ii)(c)
解法一
思路
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R R R x x x y = 0 y=0 y = 0 R R R θ \theta θ
答题过程
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At R R R y = 0 y=0 y = 0 A = 5 A=5 A = 5
5 sin ( 2 θ − 3 π 8 ) + 2 = 0 sin ( 2 θ − 3 π 8 ) = − 2 5 . \begin{align*}
5\sin\bigg(2\theta-\frac{3\pi}{8}\bigg)+2=&\,0\\
\sin\bigg(2\theta-\frac{3\pi}{8}\bigg)=&\,-\frac25.
\end{align*} 5 sin ( 2 θ − 8 3 π ) + 2 = sin ( 2 θ − 8 3 π ) = 0 − 5 2 . Let ϕ = 2 θ − 3 π 8 \phi=2\theta-\dfrac{3\pi}{8} ϕ = 2 θ − 8 3 π α = sin − 1 ( 2 / 5 ) = 0.411516846 … \alpha=\sin^{-1}(2/5)=0.411516846\ldots α = sin − 1 ( 2/5 ) = 0.411516846 …
The second maximum has ϕ = 5 π / 2 \phi=5\pi/2 ϕ = 5 π /2 R R R ϕ \phi ϕ 3 π 3\pi 3 π 7 π / 2 7\pi/2 7 π /2
ϕ = 3 π + α = 9.836294806 … . \phi=3\pi+\alpha=9.836294806\ldots. ϕ = 3 π + α = 9.836294806 … . Then
2 θ = 3 π + α + 3 π 8 θ = 3 π + α + 3 π / 8 2 = 5.507196025 … . \begin{align*}
2\theta =&\,3\pi+\alpha+\frac{3\pi}{8}\\
\theta =&\,\frac{3\pi+\alpha+3\pi/8}{2}\\
=&\,5.507196025\ldots.
\end{align*} 2 θ = θ = = 3 π + α + 8 3 π 2 3 π + α + 3 π /8 5.507196025 … . Therefore, to three significant figures,
θ = 5.51 . \boxed{\theta=5.51}. θ = 5.51 . 
