题目
Problem
(a) Express 3 3 sin ( θ + 1 6 π ) − 2 sin θ 3\sqrt3\sin\big(\theta+\frac16\pi\big)-2\sin\theta 3 3 sin ( θ + 6 1 π ) − 2 sin θ R sin ( θ + α ) R\sin(\theta+\alpha) R sin ( θ + α ) R > 0 R>0 R > 0 0 < α < 1 2 π 0<\alpha<\frac12\pi 0 < α < 2 1 π R R R α \alpha α 3 3 3
(5)
(b) Hence, solve the equation 3 3 sin ( 2 x + 1 6 π ) − 2 sin 2 x = 6 3\sqrt3\sin\big(2x+\frac16\pi\big)-2\sin2x=\sqrt6 3 3 sin ( 2 x + 6 1 π ) − 2 sin 2 x = 6 0 < x < π 0<x<\pi 0 < x < π
(4)
题目中文翻译
(a) 将 3 3 sin ( θ + 1 6 π ) − 2 sin θ 3\sqrt3\sin\big(\theta+\frac16\pi\big)-2\sin\theta 3 3 sin ( θ + 6 1 π ) − 2 sin θ R sin ( θ + α ) R\sin(\theta+\alpha) R sin ( θ + α ) R > 0 R>0 R > 0 0 < α < 1 2 π 0<\alpha<\frac12\pi 0 < α < 2 1 π R R R α \alpha α 3 3 3
(5)
(b) 由此解方程 3 3 sin ( 2 x + 1 6 π ) − 2 sin 2 x = 6 3\sqrt3\sin\big(2x+\frac16\pi\big)-2\sin2x=\sqrt6 3 3 sin ( 2 x + 6 1 π ) − 2 sin 2 x = 6 0 < x < π 0<x<\pi 0 < x < π
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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展开和角后合并正弦项，再与目标形式比较系数。振幅用平方和，锐角由正切求出。
答题过程
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Expanding the sine,
3 3 sin ( θ + π / 6 ) − 2 sin θ = 3 3 ( 3 2 sin θ + 1 2 cos θ ) − 2 sin θ = 5 2 sin θ + 3 3 2 cos θ . \begin{align*}
&\,3\sqrt3\sin(\theta+\pi/6)-2\sin\theta\\
=&\,3\sqrt3\bigg(\frac{\sqrt3}{2}\sin\theta+\frac12\cos\theta\bigg)\\[4mm]
&\,\hspace{2pt}-2\sin\theta\\[4mm]
=&\,\frac52\sin\theta+\frac{3\sqrt3}{2}\cos\theta.
\end{align*} = = 3 3 sin ( θ + π /6 ) − 2 sin θ 3 3 ( 2 3 sin θ + 2 1 cos θ ) − 2 sin θ 2 5 sin θ + 2 3 3 cos θ . Comparing with R sin ( θ + α ) R\sin(\theta+\alpha) R sin ( θ + α )
R cos α = 5 2 , R sin α = 3 3 2 . \begin{align*}
R\cos\alpha=&\,\frac52,\\[4mm]
R\sin\alpha=&\,\frac{3\sqrt3}{2}.
\end{align*} R cos α = R sin α = 2 5 , 2 3 3 . Hence
R = 25 4 + 27 4 = 13 , R=\sqrt{\frac{25}{4}+\frac{27}{4}}=\sqrt{13}, R = 4 25 + 4 27 = 13 , and
α = tan − 1 ( 3 3 5 ) = 0.804633 … = 0.805 . \alpha=\tan^{-1}\bigg(\frac{3\sqrt3}{5}\bigg)
=0.804633\ldots=\boxed{0.805}. α = tan − 1 ( 5 3 3 ) = 0.804633 … = 0.805 . The required form is 13 sin ( θ + α ) \boxed{\sqrt{13}\sin(\theta+\alpha)} 13 sin ( θ + α ) α \alpha α
(b)
解法一
思路
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把(a)的变量换成 2 x 2x 2 x
答题过程
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Using (a),
sin ( 2 x + α ) = 6 13 , \sin(2x+\alpha)=\sqrt{\frac6{13}}, sin ( 2 x + α ) = 13 6 , where α = tan − 1 ( 3 3 / 5 ) \alpha=\tan^{-1}(3\sqrt3/5) α = tan − 1 ( 3 3 /5 )
Let
β = sin − 1 6 13 = 0.746898 … . \beta=\sin^{-1}\sqrt{\frac6{13}}=0.746898\ldots. β = sin − 1 13 6 = 0.746898 … . Since 0 < x < π 0<x<\pi 0 < x < π
α < 2 x + α < 2 π + α . \alpha<2x+\alpha<2\pi+\alpha. α < 2 x + α < 2 π + α . Here β < α \beta<\alpha β < α π − β \pi-\beta π − β 2 π + β 2\pi+\beta 2 π + β
x = π − β − α 2 = 0.795030 … x=\frac{\pi-\beta-\alpha}{2}=0.795030\ldots x = 2 π − β − α = 0.795030 … or
x = 2 π + β − α 2 = 3.112725 … . x=\frac{2\pi+\beta-\alpha}{2}=3.112725\ldots. x = 2 2 π + β − α = 3.112725 … . Hence
x = 0.795 , 3.11 radians . \boxed{x=0.795,\ 3.11\text{ radians}}. x = 0.795 , 3.11 radians . 