题目
Problem
(a) By expressing 3 θ 3\theta 3 θ 2 θ + θ 2\theta+\theta 2 θ + θ cos 3 θ ≡ 4 cos 3 θ − 3 cos θ \cos3\theta\equiv4\cos^3\theta-3\cos\theta cos 3 θ ≡ 4 cos 3 θ − 3 cos θ
(3)
(b) Hence solve the equation
cos 3 θ + cos θ cos 2 θ = cos 2 θ \cos3\theta+\cos\theta\cos2\theta=\cos^2\theta cos 3 θ + cos θ cos 2 θ = cos 2 θ for 0 ∘ ≤ θ ≤ 180 ∘ 0^\circ\le\theta\le180^\circ 0 ∘ ≤ θ ≤ 18 0 ∘
(5)
题目中文翻译
(a) 将 3 θ 3\theta 3 θ 2 θ + θ 2\theta+\theta 2 θ + θ cos 3 θ ≡ 4 cos 3 θ − 3 cos θ \cos3\theta\equiv4\cos^3\theta-3\cos\theta cos 3 θ ≡ 4 cos 3 θ − 3 cos θ
(3)
(b) 由此解方程
cos 3 θ + cos θ cos 2 θ = cos 2 θ , \cos3\theta+\cos\theta\cos2\theta=\cos^2\theta, cos 3 θ + cos θ cos 2 θ = cos 2 θ , 其中 0 ∘ ≤ θ ≤ 180 ∘ 0^\circ\le\theta\le180^\circ 0 ∘ ≤ θ ≤ 18 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
(a)
解法一
思路
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按题目要求先展开 cos ( 2 θ + θ ) \cos(2\theta+\theta) cos ( 2 θ + θ ) sin 2 θ \sin^2\theta sin 2 θ 1 − cos 2 θ 1-\cos^2\theta 1 − cos 2 θ
答题过程
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Using the addition formula,
cos 3 θ = cos ( 2 θ + θ ) = cos 2 θ cos θ − sin 2 θ sin θ = ( 2 cos 2 θ − 1 ) cos θ − 2 sin 2 θ cos θ = 2 cos 3 θ − cos θ − 2 ( 1 − cos 2 θ ) cos θ = 4 cos 3 θ − 3 cos θ . \begin{align*}
\cos3\theta
=&\,\cos(2\theta+\theta)\\[4mm]
=&\,\cos2\theta\cos\theta-\sin2\theta\sin\theta\\[4mm]
=&\,(2\cos^2\theta-1)\cos\theta
-2\sin^2\theta\cos\theta\\[4mm]
=&\,2\cos^3\theta-\cos\theta
-2(1-\cos^2\theta)\cos\theta\\[4mm]
=&\,\boxed{4\cos^3\theta-3\cos\theta}.
\end{align*} cos 3 θ = = = = = cos ( 2 θ + θ ) cos 2 θ cos θ − sin 2 θ sin θ ( 2 cos 2 θ − 1 ) cos θ − 2 sin 2 θ cos θ 2 cos 3 θ − cos θ − 2 ( 1 − cos 2 θ ) cos θ 4 cos 3 θ − 3 cos θ . Hence the identity is proved.
(b)
解法一
思路
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承接(a)把方程化成 cos θ \cos\theta cos θ cos θ \cos\theta cos θ 90 ∘ 90^\circ 9 0 ∘
答题过程
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Using part (a) and cos 2 θ = 2 cos 2 θ − 1 \cos2\theta=2\cos^2\theta-1 cos 2 θ = 2 cos 2 θ − 1
4 cos 3 θ − 3 cos θ + cos θ ( 2 cos 2 θ − 1 ) = cos 2 θ . \begin{align*}
&\,4\cos^3\theta-3\cos\theta\\[4mm]
&\,\hspace{2pt}+\cos\theta(2\cos^2\theta-1)=\cos^2\theta.
\end{align*} 4 cos 3 θ − 3 cos θ + cos θ ( 2 cos 2 θ − 1 ) = cos 2 θ . Hence
cos θ ( 6 cos 2 θ − cos θ − 4 ) = 0. \cos\theta(6\cos^2\theta-\cos\theta-4)=0. cos θ ( 6 cos 2 θ − cos θ − 4 ) = 0. The first factor gives cos θ = 0 \cos\theta=0 cos θ = 0 θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘
For the quadratic factor,
cos θ = 1 ± 1 + 96 12 = 1 ± 97 12 . \cos\theta=\frac{1\pm\sqrt{1+96}}{12}
=\frac{1\pm\sqrt{97}}{12}. cos θ = 12 1 ± 1 + 96 = 12 1 ± 97 . Both values lie in [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] 0 ∘ ≤ θ ≤ 180 ∘ 0^\circ\le\theta\le180^\circ 0 ∘ ≤ θ ≤ 18 0 ∘
θ = 25.3014 … ∘ , θ = 137.5108 … ∘ . \begin{align*}
\theta=&\,25.3014\ldots^\circ,\\[4mm]
\theta=&\,137.5108\ldots^\circ.
\end{align*} θ = θ = 25.3014 … ∘ , 137.5108 … ∘ . Thus
θ = 25.3 ∘ , 90 ∘ , 137.5 ∘ . \boxed{\theta=25.3^\circ,\ 90^\circ,\ 137.5^\circ}. θ = 25. 3 ∘ , 9 0 ∘ , 137. 5 ∘ . 