题目
Problem
(a) Show that cos 4 θ − sin 4 θ − 4 sin 2 θ cos 2 θ = cos 2 2 θ + cos 2 θ − 1 \cos^4\theta-\sin^4\theta-4\sin^2\theta\cos^2\theta=\cos^22\theta+\cos2\theta-1 cos 4 θ − sin 4 θ − 4 sin 2 θ cos 2 θ = cos 2 2 θ + cos 2 θ − 1
(3)
(b) Solve the equation cos 4 α − sin 4 α = 4 sin 2 α cos 2 α \cos^4\alpha-\sin^4\alpha=4\sin^2\alpha\cos^2\alpha cos 4 α − sin 4 α = 4 sin 2 α cos 2 α 0 ∘ ≤ α ≤ 180 ∘ 0^\circ\le\alpha\le180^\circ 0 ∘ ≤ α ≤ 18 0 ∘
(3)
题目中文翻译
(a) 证明 cos 4 θ − sin 4 θ − 4 sin 2 θ cos 2 θ = cos 2 2 θ + cos 2 θ − 1 \cos^4\theta-\sin^4\theta-4\sin^2\theta\cos^2\theta=\cos^22\theta+\cos2\theta-1 cos 4 θ − sin 4 θ − 4 sin 2 θ cos 2 θ = cos 2 2 θ + cos 2 θ − 1
(3)
(b) 解方程 cos 4 α − sin 4 α = 4 sin 2 α cos 2 α \cos^4\alpha-\sin^4\alpha=4\sin^2\alpha\cos^2\alpha cos 4 α − sin 4 α = 4 sin 2 α cos 2 α 0 ∘ ≤ α ≤ 180 ∘ 0^\circ\le\alpha\le180^\circ 0 ∘ ≤ α ≤ 18 0 ∘
(3)
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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平方差先化成 cos 2 θ \cos2\theta cos 2 θ sin 2 2 θ \sin^22\theta sin 2 2 θ
答题过程
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Using the difference of squares,
cos 4 θ − sin 4 θ = ( cos 2 θ − sin 2 θ ) ( cos 2 θ + sin 2 θ ) = cos 2 θ . \begin{align*}
\cos^4\theta-\sin^4\theta
=&\,(\cos^2\theta-\sin^2\theta)
(\cos^2\theta+\sin^2\theta)\\[4mm]
=&\,\cos2\theta.
\end{align*} cos 4 θ − sin 4 θ = = ( cos 2 θ − sin 2 θ ) ( cos 2 θ + sin 2 θ ) cos 2 θ . Also 4 sin 2 θ cos 2 θ = sin 2 2 θ 4\sin^2\theta\cos^2\theta=\sin^22\theta 4 sin 2 θ cos 2 θ = sin 2 2 θ
cos 4 θ − sin 4 θ − 4 sin 2 θ cos 2 θ = cos 2 θ − sin 2 2 θ = cos 2 θ − ( 1 − cos 2 2 θ ) = cos 2 2 θ + cos 2 θ − 1 . \begin{align*}
&\,\cos^4\theta-\sin^4\theta-4\sin^2\theta\cos^2\theta\\
=&\,\cos2\theta-\sin^22\theta\\
=&\,\cos2\theta-(1-\cos^22\theta)\\
=&\,\boxed{\cos^22\theta+\cos2\theta-1}.
\end{align*} = = = cos 4 θ − sin 4 θ − 4 sin 2 θ cos 2 θ cos 2 θ − sin 2 2 θ cos 2 θ − ( 1 − cos 2 2 θ ) cos 2 2 θ + cos 2 θ − 1 .
(b)
解法一
思路
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利用(a)解关于 cos 2 α \cos2\alpha cos 2 α
答题过程
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By (a), the equation becomes
cos 2 2 α + cos 2 α − 1 = 0. \cos^22\alpha+\cos2\alpha-1=0. cos 2 2 α + cos 2 α − 1 = 0. Thus
cos 2 α = − 1 ± 5 2 . \cos2\alpha=\frac{-1\pm\sqrt5}{2}. cos 2 α = 2 − 1 ± 5 . The negative choice is less than − 1 -1 − 1
cos 2 α = 5 − 1 2 . \cos2\alpha=\frac{\sqrt5-1}{2}. cos 2 α = 2 5 − 1 . For 0 ∘ ⩽ 2 α ⩽ 360 ∘ 0^\circ\leqslant2\alpha\leqslant360^\circ 0 ∘ ⩽ 2 α ⩽ 36 0 ∘
β = cos − 1 ( 5 − 1 2 ) . \beta=\cos^{-1}\bigg(\frac{\sqrt5-1}{2}\bigg). β = cos − 1 ( 2 5 − 1 ) . Then 2 α = β 2\alpha=\beta 2 α = β 360 ∘ − β 360^\circ-\beta 36 0 ∘ − β
α = 25.9 ∘ , 154.1 ∘ . \boxed{\alpha=25.9^\circ,\ 154.1^\circ}. α = 25. 9 ∘ , 154. 1 ∘ . 