题目
Problem
(a) Show that cos 4 x + 2 sin 2 x − 1 ≡ 8 sin 4 x − 6 sin 2 x \cos4x+2\sin^2x-1\equiv8\sin^4x-6\sin^2x cos 4 x + 2 sin 2 x − 1 ≡ 8 sin 4 x − 6 sin 2 x
(4)
(b) Hence solve the equation cos 4 x + 2 sin 2 x − 1 = 0 \cos4x+2\sin^2x-1=0 cos 4 x + 2 sin 2 x − 1 = 0 − 180 ∘ ≤ x ≤ 180 ∘ -180^\circ\le x\le180^\circ − 18 0 ∘ ≤ x ≤ 18 0 ∘
(4)
题目中文翻译
(a) 证明 cos 4 x + 2 sin 2 x − 1 ≡ 8 sin 4 x − 6 sin 2 x \cos4x+2\sin^2x-1\equiv8\sin^4x-6\sin^2x cos 4 x + 2 sin 2 x − 1 ≡ 8 sin 4 x − 6 sin 2 x
(4)
(b) 由此解方程 cos 4 x + 2 sin 2 x − 1 = 0 \cos4x+2\sin^2x-1=0 cos 4 x + 2 sin 2 x − 1 = 0 − 180 ∘ ≤ x ≤ 180 ∘ -180^\circ\le x\le180^\circ − 18 0 ∘ ≤ x ≤ 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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先把四倍角余弦写成二倍角正弦，再用二倍角与平方关系把所有项化成 sin x \sin x sin x
答题过程
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Using the double-angle identities,
cos 4 x + 2 sin 2 x − 1 ≡ 1 − 2 sin 2 2 x + 2 sin 2 x − 1 ≡ − 2 ( 2 sin x cos x ) 2 + 2 sin 2 x ≡ − 8 sin 2 x ( 1 − sin 2 x ) + 2 sin 2 x ≡ − 8 sin 2 x + 8 sin 4 x + 2 sin 2 x ≡ 8 sin 4 x − 6 sin 2 x , \begin{align*}
\cos4x+2\sin^2x-1
\equiv&\,1-2\sin^22x+2\sin^2x-1\\
\equiv&\,-2(2\sin x\cos x)^2+2\sin^2x\\
\equiv&\,-8\sin^2x(1-\sin^2x)+2\sin^2x\\
\equiv&\,-8\sin^2x+8\sin^4x+2\sin^2x\\
\equiv&\,8\sin^4x-6\sin^2x,
\end{align*} cos 4 x + 2 sin 2 x − 1 ≡ ≡ ≡ ≡ ≡ 1 − 2 sin 2 2 x + 2 sin 2 x − 1 − 2 ( 2 sin x cos x ) 2 + 2 sin 2 x − 8 sin 2 x ( 1 − sin 2 x ) + 2 sin 2 x − 8 sin 2 x + 8 sin 4 x + 2 sin 2 x 8 sin 4 x − 6 sin 2 x , as required.
(b)
解法一
思路
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利用(a)因式分解。不要除以 sin 2 x \sin^2x sin 2 x
答题过程
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Using (a),
8 sin 4 x − 6 sin 2 x = 2 sin 2 x ( 4 sin 2 x − 3 ) = 0. 8\sin^4x-6\sin^2x
=2\sin^2x(4\sin^2x-3)=0. 8 sin 4 x − 6 sin 2 x = 2 sin 2 x ( 4 sin 2 x − 3 ) = 0. Thus either sin x = 0 \sin x=0 sin x = 0 sin 2 x = 3 / 4 \sin^2x=3/4 sin 2 x = 3/4
For − 180 ∘ ⩽ x ⩽ 180 ∘ -180^\circ\leqslant x\leqslant180^\circ − 18 0 ∘ ⩽ x ⩽ 18 0 ∘ sin x = 0 \sin x=0 sin x = 0
x = − 180 ∘ , 0 ∘ , 180 ∘ . x=-180^\circ,\ 0^\circ,\ 180^\circ. x = − 18 0 ∘ , 0 ∘ , 18 0 ∘ . The cases sin x = ± 3 / 2 \sin x=\pm\sqrt3/2 sin x = ± 3 /2
x = − 120 ∘ , − 60 ∘ , 60 ∘ , 120 ∘ . x=-120^\circ,\ -60^\circ,\ 60^\circ,\ 120^\circ. x = − 12 0 ∘ , − 6 0 ∘ , 6 0 ∘ , 12 0 ∘ . Hence all solutions are
x = − 180 ∘ , − 120 ∘ , − 60 ∘ , 0 ∘ , 60 ∘ , 120 ∘ , 180 ∘ . \begin{align*}
x=&\,-180^\circ,\ -120^\circ,\ -60^\circ,\ 0^\circ,\\[4mm]
&\,\hspace{2pt}60^\circ,\ 120^\circ,\ 180^\circ.
\end{align*} x = − 18 0 ∘ , − 12 0 ∘ , − 6 0 ∘ , 0 ∘ , 6 0 ∘ , 12 0 ∘ , 18 0 ∘ . 