题目
Problem
(a) Express 3 cos x + 2 cos ( x − 60 ∘ ) 3\cos x+2\cos(x-60^\circ) 3 cos x + 2 cos ( x − 6 0 ∘ ) R cos ( x − α ) R\cos(x-\alpha) R cos ( x − α ) R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘ R R R α \alpha α 2 2 2
(4)
(b) Hence solve the equation
3 cos 2 θ + 2 cos ( 2 θ − 60 ∘ ) = 2.5 3\cos2\theta+2\cos(2\theta-60^\circ)=2.5 3 cos 2 θ + 2 cos ( 2 θ − 6 0 ∘ ) = 2.5 for 0 ∘ < θ < 180 ∘ 0^\circ<\theta<180^\circ 0 ∘ < θ < 18 0 ∘
(4)
题目中文翻译
(a) 将 3 cos x + 2 cos ( x − 60 ∘ ) 3\cos x+2\cos(x-60^\circ) 3 cos x + 2 cos ( x − 6 0 ∘ ) R cos ( x − α ) R\cos(x-\alpha) R cos ( x − α ) R > 0 R>0 R > 0 0 ∘ < α < 90 ∘ 0^\circ<\alpha<90^\circ 0 ∘ < α < 9 0 ∘ R R R α \alpha α 2 2 2
(4)
(b) 由此解方程
3 cos 2 θ + 2 cos ( 2 θ − 60 ∘ ) = 2.5 , 3\cos2\theta+2\cos(2\theta-60^\circ)=2.5, 3 cos 2 θ + 2 cos ( 2 θ − 6 0 ∘ ) = 2.5 , 其中 0 ∘ < θ < 180 ∘ 0^\circ<\theta<180^\circ 0 ∘ < θ < 18 0 ∘
(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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先展开 cos ( x − 60 ∘ ) \cos(x-60^\circ) cos ( x − 6 0 ∘ ) R cos ( x − α ) R\cos(x-\alpha) R cos ( x − α )
答题过程
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Using the addition formula,
3 cos x + 2 cos ( x − 60 ∘ ) = 3 cos x + 2 ( 1 2 cos x + 3 2 sin x ) = 4 cos x + 3 sin x . \begin{align*}
&\,3\cos x+2\cos(x-60^\circ)\\[4mm]
=&\,3\cos x\\[4mm]
&\,\hspace{2pt}+2\bigg(\frac12\cos x+\frac{\sqrt3}{2}\sin x\bigg)\\[4mm]
=&\,4\cos x+\sqrt3\sin x.
\end{align*} = = 3 cos x + 2 cos ( x − 6 0 ∘ ) 3 cos x + 2 ( 2 1 cos x + 2 3 sin x ) 4 cos x + 3 sin x . Also,
R cos ( x − α ) = R cos α cos x + R sin α sin x . R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x. R cos ( x − α ) = R cos α cos x + R sin α sin x . Thus
R cos α = 4 , R sin α = 3 . \begin{align*}
R\cos\alpha=&\,4,\\[4mm]
R\sin\alpha=&\,\sqrt3.
\end{align*} R cos α = R sin α = 4 , 3 . Hence
R = 16 + 3 = 19 , α = tan − 1 3 4 = 23.413224 … ∘ . \begin{align*}
R=&\,\sqrt{16+3}=\sqrt{19},\\[4mm]
\alpha=&\,\tan^{-1}\frac{\sqrt3}{4}\\[4mm]
=&\,23.413224\ldots^\circ.
\end{align*} R = α = = 16 + 3 = 19 , tan − 1 4 3 23.413224 … ∘ . Therefore
R = 19 , α ≈ 23.41 ∘ . R=\boxed{\sqrt{19}},\quad\alpha\approx\boxed{23.41^\circ}. R = 19 , α ≈ 23.4 1 ∘ .
(b)
解法一
思路
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把(a)中的 x x x 2 θ 2\theta 2 θ 2 2 2
答题过程
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Using part (a), with the unrounded value of α \alpha α
19 cos ( 2 θ − α ) = 2.5. \sqrt{19}\cos(2\theta-\alpha)=2.5. 19 cos ( 2 θ − α ) = 2.5. Let
β = cos − 1 2.5 19 = 55.002595 … ∘ . \beta=\cos^{-1}\frac{2.5}{\sqrt{19}}
=55.002595\ldots^\circ. β = cos − 1 19 2.5 = 55.002595 … ∘ . Since 0 ∘ < 2 θ < 360 ∘ 0^\circ<2\theta<360^\circ 0 ∘ < 2 θ < 36 0 ∘
2 θ = α + β or 2 θ = α + 360 ∘ − β . \begin{align*}
2\theta=&\,\alpha+\beta\\[4mm]
\text{or}\quad2\theta=&\,\alpha+360^\circ-\beta.
\end{align*} 2 θ = or 2 θ = α + β α + 36 0 ∘ − β . Therefore
θ = 39.2 ∘ or 164.2 ∘ . \boxed{\theta=39.2^\circ\quad\text{or}\quad164.2^\circ}. θ = 39. 2 ∘ or 164. 2 ∘ . 