在坐標平面上,廣義角 \(\theta\) 的頂點為原點 \(O\),始邊為 \(x\) 軸的正向,且滿足 \(\tan \theta = \frac{2}{3}\)。若 \(\theta\) 的終邊上有一點 \(P\),其 \(y\) 坐標為 \(-4\),則下列哪些選項一定正確？
(1) \(P\) 的 \(x\) 坐標是 6
(2) \(OP = 2\sqrt{13}\)
(3) \(\cos \theta = \frac{3}{\sqrt{13}}\)
(4) \(\sin 2\theta > 0\)
(5) \(\cos \frac{\theta}{2} < 0\)
\[
\boxed{\text{已知條件與初步計算}}
\]
\[
\begin{aligned}
& \text{已知點 } P(x, -4) \text{ 在角 } \theta \text{ 的終邊上，且 } \tan\theta = \frac{2}{3} \\
\\
& \text{由定義：} \tan\theta = \frac{y}{x} = \frac{-4}{x} = \frac{2}{3} \\
& \Rightarrow \frac{-4}{x} = \frac{2}{3} \quad \Rightarrow \quad 2x = -12 \quad \Rightarrow \quad x = -6 \\
\\
& \therefore P \text{ 點座標為 } (-6, -4)，\theta \text{ 為第三象限角}
\end{aligned}
\]
\[
\boxed{\text{各選項分析與判斷}}
\]
\[
\begin{aligned}
\text{(1) } & \tan\theta = \frac{2}{3} \Rightarrow x = -6 \\
& \text{題目中已直接計算得到，並非「根據定義」立即可得 } x = -6 \\
& \text{實際需經過計算：} \frac{-4}{x} = \frac{2}{3} \Rightarrow x = -6 \\
& \therefore \text{此選項描述不準確} \quad (\times)
\end{aligned}
\]
\[
\begin{aligned}
\text{(2) } & \overline{OP} = \sqrt{x^2 + y^2} = \sqrt{(-6)^2 + (-4)^2} \\
& = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \\
& \therefore \text{正確} \quad (\bigcirc)
\end{aligned}
\]
\[
\begin{aligned}
\text{(3) } & \cos\theta = \frac{x}{\overline{OP}} = \frac{-6}{2\sqrt{13}} = -\frac{3}{\sqrt{13}} \\
& \text{但選項寫為 } \frac{3}{\sqrt{13}} \text{（缺少負號）} \\
& \therefore \text{錯誤} \quad (\times)
\end{aligned}
\]
\[
\begin{aligned}
\text{(4) } & \sin 2\theta = 2\sin\theta\cos\theta \\
& \text{其中 } \sin\theta = \frac{y}{\overline{OP}} = \frac{-4}{2\sqrt{13}} = -\frac{2}{\sqrt{13}} \\
& \cos\theta = -\frac{3}{\sqrt{13}} \\
& \sin 2\theta = 2 \times \left(-\frac{2}{\sqrt{13}}\right) \times \left(-\frac{3}{\sqrt{13}}\right) \\
& = 2 \times \frac{6}{13} = \frac{12}{13} > 0 \\
& \therefore \text{正確} \quad (\bigcirc)
\end{aligned}
\]
\[
\boxed{\text{選項(5)詳細分析：半角象限判斷}}
\]
\[
\begin{aligned}
& \text{已知 } \theta \text{ 為第三象限角：} \\
& 180^\circ + 360^\circ n < \theta < 270^\circ + 360^\circ n \quad (n \in \mathbb{Z}) \\
\\
& \text{不等式各項除以 2：} \\
& 90^\circ + 180^\circ n < \frac{\theta}{2} < 135^\circ + 180^\circ n \\
\\
& \text{分情況討論：} \\
& \bullet \text{當 } n \text{ 為偶數（設 } n=2k\text{）：} \\
& \quad 90^\circ + 360^\circ k < \frac{\theta}{2} < 135^\circ + 360^\circ k \\
& \quad \Rightarrow \frac{\theta}{2} \text{ 為第二象限角} \Rightarrow \cos\frac{\theta}{2} < 0 \\
\\
& \bullet \text{當 } n \text{ 為奇數（設 } n=2k+1\text{）：} \\
& \quad 90^\circ + 180^\circ(2k+1) < \frac{\theta}{2} < 135^\circ + 180^\circ(2k+1) \\
& \quad = 270^\circ + 360^\circ k < \frac{\theta}{2} < 315^\circ + 360^\circ k \\
& \quad \Rightarrow \frac{\theta}{2} \text{ 為第四象限角} \Rightarrow \cos\frac{\theta}{2} > 0 \\
\\
& \therefore \cos\frac{\theta}{2} \text{ 可能為正也可能為負} \\
& \text{選項(5)斷言其必為正或負是錯誤的} \quad (\times)
\end{aligned}
\]
\[
\boxed{\text{綜合結論}}
\]
\[
\begin{array}{c|c|c}
\text{選項} & \text{內容} & \text{判斷} \\ \hline
(1) & x = -6 \text{ 是根據定義直接可得} & \times \\
(2) & \overline{OP} = 2\sqrt{13} & \bigcirc \\
(3) & \cos\theta = \dfrac{3}{\sqrt{13}} & \times \\
(4) & \sin 2\theta = \dfrac{12}{13} > 0 & \bigcirc \\
(5) & \cos\dfrac{\theta}{2} \text{ 必為正（或負）} & \times \\
\end{array}
\]
\[
\boxed{\text{故正確答案為： (2)(4)}}
\]
\[
\boxed{\text{※ 三角函數半角象限判斷通則}}
\]
\[
\begin{aligned}
\text{若 } \theta \text{ 在第 } k \text{ 象限，則 } \frac{\theta}{2} \text{ 的象限分佈：} \\
\begin{array}{c|c}
\theta \text{ 所在象限} & \dfrac{\theta}{2} \text{ 可能象限} \\ \hline
\text{第一象限} & \text{第一、三象限} \\
\text{第二象限} & \text{第一、三象限} \\
\text{第三象限} & \text{第二、四象限} \\
\text{第四象限} & \text{第二、四象限} \\
\end{array}
\end{aligned}
\]
\[
\boxed{\text{本題關鍵計算回顧}}
\]
\[
\begin{aligned}
P(-6, -4) &\Rightarrow \overline{OP} = 2\sqrt{13} \\
\sin\theta &= -\frac{2}{\sqrt{13}}, \quad \cos\theta = -\frac{3}{\sqrt{13}} \\
\sin 2\theta &= 2\left(-\frac{2}{\sqrt{13}}\right)\left(-\frac{3}{\sqrt{13}}\right) = \frac{12}{13}
\end{aligned}
\]
