题目
Problem
(a) Prove the identity cot 2 θ − tan 2 θ ≡ 4 cot 2 θ cosec 2 θ \cot^2\theta-\tan^2\theta\equiv4\cot2\theta\operatorname{cosec}2\theta cot 2 θ − tan 2 θ ≡ 4 cot 2 θ cosec 2 θ
(4)
(b) Hence solve the equation cot 2 x − tan 2 x = 5 sec 2 x \cot^2x-\tan^2x=5\sec2x cot 2 x − tan 2 x = 5 sec 2 x 0 ∘ < x < 90 ∘ 0^\circ<x<90^\circ 0 ∘ < x < 9 0 ∘
(4)
题目中文翻译
(a) 证明恒等式 cot 2 θ − tan 2 θ ≡ 4 cot 2 θ cosec 2 θ \cot^2\theta-\tan^2\theta\equiv4\cot2\theta\operatorname{cosec}2\theta cot 2 θ − tan 2 θ ≡ 4 cot 2 θ cosec 2 θ
(4)
(b) 由此解方程 cot 2 x − tan 2 x = 5 sec 2 x \cot^2x-\tan^2x=5\sec2x cot 2 x − tan 2 x = 5 sec 2 x 0 ∘ < x < 90 ∘ 0^\circ<x<90^\circ 0 ∘ < x < 9 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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通分后把分子作为平方差分解，再利用二倍角化简分子与分母。
答题过程
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Where both sides are defined,
cot 2 θ − tan 2 θ ≡ cos 2 θ sin 2 θ − sin 2 θ cos 2 θ ≡ cos 4 θ − sin 4 θ sin 2 θ cos 2 θ ≡ ( cos 2 θ − sin 2 θ ) ( cos 2 θ + sin 2 θ ) sin 2 θ cos 2 θ ≡ 4 cos 2 θ sin 2 2 θ ≡ 4 cot 2 θ cosec 2 θ . \begin{align*}
\cot^2\theta-\tan^2\theta
\equiv&\,\frac{\cos^2\theta}{\sin^2\theta}
-\frac{\sin^2\theta}{\cos^2\theta}\\
\equiv&\,\frac{\cos^4\theta-\sin^4\theta}
{\sin^2\theta\cos^2\theta}\\
\equiv&\,\frac{(\cos^2\theta-\sin^2\theta)
(\cos^2\theta+\sin^2\theta)}
{\sin^2\theta\cos^2\theta}\\
\equiv&\,\frac{4\cos2\theta}{\sin^22\theta}\\
\equiv&\,4\cot2\theta\operatorname{cosec}2\theta.
\end{align*} cot 2 θ − tan 2 θ ≡ ≡ ≡ ≡ ≡ sin 2 θ cos 2 θ − cos 2 θ sin 2 θ sin 2 θ cos 2 θ cos 4 θ − sin 4 θ sin 2 θ cos 2 θ ( cos 2 θ − sin 2 θ ) ( cos 2 θ + sin 2 θ ) sin 2 2 θ 4 cos 2 θ 4 cot 2 θ cosec 2 θ .
解法二
思路
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也可直接分解成 ( cot θ − tan θ ) ( cot θ + tan θ ) (\cot\theta-\tan\theta)(\cot\theta+\tan\theta) ( cot θ − tan θ ) ( cot θ + tan θ )
答题过程
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The two factors satisfy
cot θ − tan θ = cos 2 θ − sin 2 θ sin θ cos θ = 2 cot 2 θ , \cot\theta-\tan\theta
=\frac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}
=2\cot2\theta, cot θ − tan θ = sin θ cos θ cos 2 θ − sin 2 θ = 2 cot 2 θ , and
cot θ + tan θ = cos 2 θ + sin 2 θ sin θ cos θ = 2 cosec 2 θ . \cot\theta+\tan\theta
=\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}
=2\operatorname{cosec}2\theta. cot θ + tan θ = sin θ cos θ cos 2 θ + sin 2 θ = 2 cosec 2 θ . Multiplying,
cot 2 θ − tan 2 θ ≡ ( cot θ − tan θ ) ( cot θ + tan θ ) ≡ 4 cot 2 θ cosec 2 θ , \begin{align*}
\cot^2\theta-\tan^2\theta
\equiv&\,(\cot\theta-\tan\theta)(\cot\theta+\tan\theta)\\
\equiv&\,4\cot2\theta\operatorname{cosec}2\theta,
\end{align*} cot 2 θ − tan 2 θ ≡ ≡ ( cot θ − tan θ ) ( cot θ + tan θ ) 4 cot 2 θ cosec 2 θ , as required.
解法三
思路
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先用勾股恒等式把余切平方、正切平方分别改写为余割平方、正割平方，常数项抵消后直接通分。
答题过程
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Where the expressions are defined, use cot 2 θ = cosec 2 θ − 1 \cot^2\theta=\operatorname{cosec}^2\theta-1 cot 2 θ = cosec 2 θ − 1 tan 2 θ = sec 2 θ − 1 \tan^2\theta=\sec^2\theta-1 tan 2 θ = sec 2 θ − 1
cot 2 θ − tan 2 θ ≡ cosec 2 θ − sec 2 θ ≡ cos 2 θ − sin 2 θ sin 2 θ cos 2 θ ≡ 4 cos 2 θ sin 2 2 θ ≡ 4 cot 2 θ cosec 2 θ . \begin{align*}
\cot^2\theta-\tan^2\theta
\equiv&\,\operatorname{cosec}^2\theta-\sec^2\theta\\[4mm]
\equiv&\,\frac{\cos^2\theta-\sin^2\theta}
{\sin^2\theta\cos^2\theta}\\[4mm]
\equiv&\,\frac{4\cos2\theta}{\sin^22\theta}\\[4mm]
\equiv&\,4\cot2\theta\operatorname{cosec}2\theta.
\end{align*} cot 2 θ − tan 2 θ ≡ ≡ ≡ ≡ cosec 2 θ − sec 2 θ sin 2 θ cos 2 θ cos 2 θ − sin 2 θ sin 2 2 θ 4 cos 2 θ 4 cot 2 θ cosec 2 θ .
(b)
解法一
思路
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利用(a)后乘去非零分母，得到关于 tan 2 x \tan2x tan 2 x x = 45 ∘ x=45^\circ x = 4 5 ∘
答题过程
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For 0 ∘ < x < 90 ∘ 0^\circ<x<90^\circ 0 ∘ < x < 9 0 ∘ x = 45 ∘ x=45^\circ x = 4 5 ∘ sec 2 x \sec2x sec 2 x
Using (a),
4 cos 2 x sin 2 2 x = 5 cos 2 x . \frac{4\cos2x}{\sin^22x}=\frac5{\cos2x}. sin 2 2 x 4 cos 2 x = cos 2 x 5 . Multiplying by the nonzero denominators,
4 cos 2 2 x = 5 sin 2 2 x . 4\cos^22x=5\sin^22x. 4 cos 2 2 x = 5 sin 2 2 x . Hence
tan 2 2 x = 4 5 , tan 2 x = ± 2 5 . \tan^22x=\frac45,
\qquad
\tan2x=\pm\frac2{\sqrt5}. tan 2 2 x = 5 4 , tan 2 x = ± 5 2 . Since 0 ∘ < 2 x < 180 ∘ 0^\circ<2x<180^\circ 0 ∘ < 2 x < 18 0 ∘ β = tan − 1 ( 2 / 5 ) \beta=\tan^{-1}(2/\sqrt5) β = tan − 1 ( 2/ 5 ) 2 x = β 2x=\beta 2 x = β 2 x = 180 ∘ − β 2x=180^\circ-\beta 2 x = 18 0 ∘ − β
Thus
x = 20.9 ∘ , 69.1 ∘ . \boxed{x=20.9^\circ,\ 69.1^\circ}. x = 20. 9 ∘ , 69. 1 ∘ . Both are valid in the original equation.
