题目
Problem
(a) Show that
3 tan 2 θ + 5 sin 2 θ = 8 sin 2 θ − 5 sin 4 θ 1 − sin 2 θ . 3\tan^2\theta + 5\sin^2\theta = \frac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta}. 3 tan 2 θ + 5 sin 2 θ = 1 − sin 2 θ 8 sin 2 θ − 5 sin 4 θ . [3]
(b) Hence solve the equation 3 tan 2 θ + 5 sin 2 θ = 9 3\tan^2\theta + 5\sin^2\theta = 9 3 tan 2 θ + 5 sin 2 θ = 9 0 ∘ < θ < 270 ∘ 0^\circ < \theta < 270^\circ 0 ∘ < θ < 27 0 ∘
[4]
题目中文翻译
(a) 证明
3 tan 2 θ + 5 sin 2 θ = 8 sin 2 θ − 5 sin 4 θ 1 − sin 2 θ 3\tan^2\theta + 5\sin^2\theta = \frac{8\sin^2\theta - 5\sin^4\theta}{1 - \sin^2\theta} 3 tan 2 θ + 5 sin 2 θ = 1 − sin 2 θ 8 sin 2 θ − 5 sin 4 θ [3]
(b) 由此解方程 3 tan 2 θ + 5 sin 2 θ = 9 3\tan^2\theta + 5\sin^2\theta = 9 3 tan 2 θ + 5 sin 2 θ = 9 0 ∘ < θ < 270 ∘ 0^\circ < \theta < 270^\circ 0 ∘ < θ < 27 0 ∘
[4]
解答
(a)
解法一
思路
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把 tan 2 θ \tan^2\theta tan 2 θ sin 2 θ / cos 2 θ \sin^2\theta/\cos^2\theta sin 2 θ / cos 2 θ
答题过程
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When cos θ ≠ 0 \cos\theta\ne0 cos θ = 0
3 tan 2 θ + 5 sin 2 θ = 3 sin 2 θ 1 − sin 2 θ + 5 sin 2 θ = 3 sin 2 θ + 5 sin 2 θ ( 1 − sin 2 θ ) 1 − sin 2 θ = 8 sin 2 θ − 5 sin 4 θ 1 − sin 2 θ . \begin{align*}
&\,3\tan^2\theta+5\sin^2\theta\\
=&\,\frac{3\sin^2\theta}{1-\sin^2\theta}+5\sin^2\theta\\
=&\,\frac{3\sin^2\theta+5\sin^2\theta(1-\sin^2\theta)}{1-\sin^2\theta}\\
=&\,\frac{8\sin^2\theta-5\sin^4\theta}{1-\sin^2\theta}.
\end{align*} = = = 3 tan 2 θ + 5 sin 2 θ 1 − sin 2 θ 3 sin 2 θ + 5 sin 2 θ 1 − sin 2 θ 3 sin 2 θ + 5 sin 2 θ ( 1 − sin 2 θ ) 1 − sin 2 θ 8 sin 2 θ − 5 sin 4 θ .
(b)
解法一
思路
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承接 (a)，令 u = sin 2 θ u=\sin^2\theta u = sin 2 θ
答题过程
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From (a), 8 u − 5 u 2 = 9 ( 1 − u ) 8u-5u^2=9(1-u) 8 u − 5 u 2 = 9 ( 1 − u )
5 u 2 − 17 u + 9 = 0 , u = 17 ± 109 10 . 5u^2-17u+9=0,
\qquad u=\frac{17\pm\sqrt{109}}{10}. 5 u 2 − 17 u + 9 = 0 , u = 10 17 ± 109 . The larger root exceeds 1 1 1 u = ( 17 − 109 ) / 10 u=(17-\sqrt{109})/10 u = ( 17 − 109 ) /10 sin θ = ± u \sin\theta=\pm\sqrt u sin θ = ± u α = sin − 1 ( u ) ≈ 54.1 ∘ \alpha=\sin^{-1}(\sqrt u)\approx54.1^\circ α = sin − 1 ( u ) ≈ 54. 1 ∘ 0 ∘ < θ < 270 ∘ 0^\circ<\theta<270^\circ 0 ∘ < θ < 27 0 ∘ α \alpha α 180 ∘ − α 180^\circ-\alpha 18 0 ∘ − α 180 ∘ + α 180^\circ+\alpha 18 0 ∘ + α 54.1 ∘ 54.1^\circ 54. 1 ∘ 125.9 ∘ 125.9^\circ 125. 9 ∘ 234.1 ∘ 234.1^\circ 234. 1 ∘
