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CIE 9709 2025 June Paper 12 Q7

A Level / CIE / P1

CIE 9709 2025 June Paper 12 Paper · Question 7

题目

Problem

(a) Prove the identity

tan⁡θ+7tan⁡2θ−3≡sin⁡θcos⁡θ+7cos⁡2θ1−4cos⁡2θ.\frac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}.
[3]

(b) Hence solve the equation

sin⁡θcos⁡θ+7cos⁡2θ1−4cos⁡2θ=5tan⁡θ\frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta} = \frac{5}{\tan\theta}

for 0∘≤θ≤180∘0^\circ \leq \theta \leq 180^\circ.

[4]
题目中文翻译

(a) 证明恒等式

tan⁡θ+7tan⁡2θ−3≡sin⁡θcos⁡θ+7cos⁡2θ1−4cos⁡2θ\frac{\tan\theta + 7}{\tan^2\theta - 3} \equiv \frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta}
[3]

(b) 由此解方程

sin⁡θcos⁡θ+7cos⁡2θ1−4cos⁡2θ=5tan⁡θ\frac{\sin\theta\cos\theta + 7\cos^2\theta}{1 - 4\cos^2\theta} = \frac{5}{\tan\theta}

其中 0∘≤θ≤180∘0^\circ \leq \theta \leq 180^\circ。

[4]

解答

(a)

解法一

思路

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从左式开始，把正切写成正弦除以余弦。分子和分母同乘 cos⁡2θ\cos^2\theta 后，分母再用平方恒等式化简。

答题过程

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On the domain of the left side, cos⁡θ≠0\cos\theta\ne0 and tan⁡2θ≠3\tan^2\theta\ne3. Thus

 tan⁡θ+7tan⁡2θ−3= sin⁡θcos⁡θ+7cos⁡2θsin⁡2θ−3cos⁡2θ= sin⁡θcos⁡θ+7cos⁡2θ1−4cos⁡2θ.\begin{align*} &\,\frac{\tan\theta+7}{\tan^2\theta-3}\\ =&\,\frac{\sin\theta\cos\theta+7\cos^2\theta}{\sin^2\theta-3\cos^2\theta}\\ =&\,\frac{\sin\theta\cos\theta+7\cos^2\theta}{1-4\cos^2\theta}. \end{align*}

(b)

解法一

思路

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用 (a) 把左边换成正切的式子。先排除使原式无定义的角，再通分并求解二次方程。

答题过程

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Let t=tan⁡θt=\tan\theta. The original equation requires tt to be defined, t≠0t\ne0 and t2≠3t^2\ne3. Using (a) gives

t+7t2−3=5t⟹4t2−7t−15=(4t+5)(t−3)=0.\frac{t+7}{t^2-3}=\frac5t \quad\Longrightarrow\quad 4t^2-7t-15=(4t+5)(t-3)=0.

Thus t=3t=3 or t=−54t=-\frac54, both allowed. In the required range these give θ=tan⁡−1(3)≈71.6∘\theta=\tan^{-1}(3)\approx71.6^\circ and θ=180∘−tan⁡−1(5/4)≈128.7∘\theta=180^\circ-\tan^{-1}(5/4)\approx128.7^\circ.