题目
Problem
Solve the equation cos ( θ − 60 ∘ ) = 3 sin θ \cos(\theta-60^\circ)=3\sin\theta cos ( θ − 6 0 ∘ ) = 3 sin θ 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ\leq\theta\leq360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
(5)
题目中文翻译
解方程 cos ( θ − 60 ∘ ) = 3 sin θ \cos(\theta-60^\circ)=3\sin\theta cos ( θ − 6 0 ∘ ) = 3 sin θ 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ\leq\theta\leq360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
(5)
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
解法一
思路
展开
展开余弦差角，用特殊角的精确值化简，再求正切。正切为正，解在第一、第三象限。
答题过程
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Expanding,
cos θ cos 60 ∘ + sin θ sin 60 ∘ = 3 sin θ . \cos\theta\cos60^\circ+\sin\theta\sin60^\circ
=3\sin\theta. cos θ cos 6 0 ∘ + sin θ sin 6 0 ∘ = 3 sin θ . Thus
1 2 cos θ + 3 2 sin θ = 3 sin θ , \frac12\cos\theta+\frac{\sqrt3}{2}\sin\theta
=3\sin\theta, 2 1 cos θ + 2 3 sin θ = 3 sin θ , so
cos θ = ( 6 − 3 ) sin θ . \cos\theta=(6-\sqrt3)\sin\theta. cos θ = ( 6 − 3 ) sin θ . Here cos θ = 0 \cos\theta=0 cos θ = 0
tan θ = 1 6 − 3 . \tan\theta=\frac1{6-\sqrt3}. tan θ = 6 − 3 1 . The first-quadrant angle is 13.1867 … ∘ 13.1867\ldots^\circ 13.1867 … ∘ 180 ∘ 180^\circ 18 0 ∘
θ = 13.2 ∘ or 193.2 ∘ . \boxed{\theta=13.2^\circ\quad\text{or}\quad193.2^\circ}. θ = 13. 2 ∘ or 193. 2 ∘ .
解法二
思路
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展开后也可以把正弦、余弦的线性组合写成辅助角形式，再解余弦为零的方程。
答题过程
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Expanding the original equation and moving all terms to the left,
1 2 cos θ − ( 3 − 3 2 ) sin θ = 0. \frac12\cos\theta-\bigg(3-\frac{\sqrt3}{2}\bigg)\sin\theta=0. 2 1 cos θ − ( 3 − 2 3 ) sin θ = 0. Write the left side as R cos ( θ + α ) R\cos(\theta+\alpha) R cos ( θ + α )
R cos α = 1 2 , R sin α = 3 − 3 2 . \begin{align*}
R\cos\alpha=&\,\frac12,\\[4mm]
R\sin\alpha=&\,3-\frac{\sqrt3}{2}.
\end{align*} R cos α = R sin α = 2 1 , 3 − 2 3 . Hence
α = tan − 1 ( 6 − 3 ) = 76.8132 … ∘ . \alpha=\tan^{-1}(6-\sqrt3)=76.8132\ldots^\circ. α = tan − 1 ( 6 − 3 ) = 76.8132 … ∘ . The equation is therefore cos ( θ + α ) = 0 \cos(\theta+\alpha)=0 cos ( θ + α ) = 0
θ + α = 90 ∘ or 270 ∘ . \theta+\alpha=90^\circ\quad\text{or}\quad270^\circ. θ + α = 9 0 ∘ or 27 0 ∘ . Thus
θ = 13.2 ∘ or 193.2 ∘ . \boxed{\theta=13.2^\circ\quad\text{or}\quad193.2^\circ}. θ = 13. 2 ∘ or 193. 2 ∘ . 