## 1. 問題の内容

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1. 問題の内容

画像に記載された数学の問題を解きます。具体的には以下の7問です。

1. 極限 $\lim_{x \to \frac{\pi}{6}} \frac{\sin x - \frac{1}{2}}{x - \frac{\pi}{6}}$ を求める。

2. 複素数平面上の3点O(0), A(2+$\sqrt{3}i$), Bがあり、$\angle AOB = \frac{\pi}{6}$ かつ OA = 2OBを満たしているとき、点Bを表す複素数を求める。

3. $i$を虚数単位とするとき、$(\frac{1+\sqrt{3}i}{1+i})^{12}=a+bi$を満たす実数$a, b$を求める。

4. 複素数$\frac{5-2i}{7+3i}$の偏角を$\theta$とするとき、$\theta$を求める（ただし、$i=\sqrt{-1}$, $0\le\theta<2\pi$）。

5. $z=1+2i$のとき、$z^5$を求める。また、複素数平面上の3点O(0), A(z), B($z^3$)に対して, $\sin \angle AOB$の値を求める（ただし、$i=\sqrt{-1}$）。

6. $\alpha=1+i, \beta=(1-\sqrt{2})+(1+\sqrt{2})i$とし、複素数平面上の3点O(0), A($\alpha$), B($\beta$)を考える。このとき、$\frac{\beta}{\alpha}$を計算し、三角形OABの面積を求める（ただし、$i=\sqrt{-1}$）。

7. 等比数列 $\{a_n\}$ が $a_1 = \sqrt{3}-1, a_2 = 4-2\sqrt{3}$ を満たすとき、無限級数 $\sum_{n=1}^{\infty} a_n$ の和を求める。

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2. 解き方の手順

1. **極限**

* ロピタルの定理を用いる。$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

* $f(x) = \sin x - \frac{1}{2}$, $g(x) = x - \frac{\pi}{6}$

* $f'(x) = \cos x$, $g'(x) = 1$

* $\lim_{x \to \frac{\pi}{6}} \frac{\cos x}{1} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

2. **複素数平面**

* $OA = |2 + \sqrt{3}i| = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4+3} = \sqrt{7}$

* $OB = \frac{OA}{2} = \frac{\sqrt{7}}{2}$

* 点Bを表す複素数を$z$とおくと、$z = OB(\cos(\frac{\pi}{6}+\theta) + i \sin(\frac{\pi}{6}+\theta))$。ここで$\theta$はOAが実軸となす角であり、A($2+\sqrt{3}i$)より$\cos\theta = \frac{2}{\sqrt{7}}, \sin \theta = \frac{\sqrt{3}}{\sqrt{7}}$。

* $\cos(\frac{\pi}{6}+\theta) = \cos\frac{\pi}{6}\cos\theta - \sin\frac{\pi}{6}\sin\theta = \frac{\sqrt{3}}{2}\frac{2}{\sqrt{7}} - \frac{1}{2}\frac{\sqrt{3}}{\sqrt{7}} = \frac{\sqrt{3}}{2\sqrt{7}}$

* $\sin(\frac{\pi}{6}+\theta) = \sin\frac{\pi}{6}\cos\theta + \cos\frac{\pi}{6}\sin\theta = \frac{1}{2}\frac{2}{\sqrt{7}} + \frac{\sqrt{3}}{2}\frac{\sqrt{3}}{\sqrt{7}} = \frac{2+3}{2\sqrt{7}} = \frac{5}{2\sqrt{7}}$

* $z = \frac{\sqrt{7}}{2}(\frac{\sqrt{3}}{2\sqrt{7}} + i\frac{5}{2\sqrt{7}}) = \frac{\sqrt{3}}{4} + i\frac{5}{4}$

3. **複素数計算**

* $\frac{1+\sqrt{3}i}{1+i} = \frac{(1+\sqrt{3}i)(1-i)}{(1+i)(1-i)} = \frac{1 - i + \sqrt{3}i - \sqrt{3}i^2}{1-i^2} = \frac{1 - i + \sqrt{3}i + \sqrt{3}}{2} = \frac{1+\sqrt{3}}{2} + \frac{\sqrt{3}-1}{2}i$

* $z = \frac{1+\sqrt{3}}{2} + \frac{\sqrt{3}-1}{2}i = r (\cos\theta + i\sin\theta)$とおく

* $r = \sqrt{(\frac{1+\sqrt{3}}{2})^2 + (\frac{\sqrt{3}-1}{2})^2} = \sqrt{\frac{1 + 2\sqrt{3} + 3 + 3 - 2\sqrt{3} + 1}{4}} = \sqrt{\frac{8}{4}} = \sqrt{2}$

* $\theta = \arctan(\frac{\sqrt{3}-1}{1+\sqrt{3}}) = \arctan(\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{3-1}) = \arctan(\frac{3-2\sqrt{3}+1}{2}) = \arctan(2-\sqrt{3})=\frac{\pi}{12}$

* $z = \sqrt{2}(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12})$

* $z^{12} = (\sqrt{2})^{12} (\cos\pi + i\sin\pi) = 2^6 (-1 + 0i) = -64$

* したがって、$a = -64, b = 0$

4. **複素数の偏角**

* $\frac{5-2i}{7+3i} = \frac{(5-2i)(7-3i)}{(7+3i)(7-3i)} = \frac{35 - 15i - 14i + 6i^2}{49 - 9i^2} = \frac{35 - 29i - 6}{49 + 9} = \frac{29 - 29i}{58} = \frac{1}{2} - \frac{1}{2}i$

* 偏角$\theta$は$\tan\theta = \frac{-\frac{1}{2}}{\frac{1}{2}} = -1$ を満たす。$0 \le \theta < 2\pi$より、$\theta = \frac{7\pi}{4}$

5. **複素数計算とsin**

* $z = 1 + 2i$

* $z^2 = (1+2i)^2 = 1 + 4i + 4i^2 = 1 + 4i - 4 = -3 + 4i$

* $z^3 = z^2 \cdot z = (-3+4i)(1+2i) = -3 - 6i + 4i + 8i^2 = -3 - 2i - 8 = -11 - 2i$

* $z^5 = z^2 \cdot z^3 = (-3+4i)(-11-2i) = 33 + 6i - 44i - 8i^2 = 33 - 38i + 8 = 41 - 38i$

* $\overrightarrow{OA} = (1,2)$, $\overrightarrow{OB} = (-11, -2)$

* $\cos \angle AOB = \frac{\overrightarrow{OA} \cdot \overrightarrow{OB}}{|\overrightarrow{OA}||\overrightarrow{OB}|} = \frac{(1)(-11) + (2)(-2)}{\sqrt{1^2+2^2} \sqrt{(-11)^2+(-2)^2}} = \frac{-11-4}{\sqrt{5}\sqrt{125}} = \frac{-15}{\sqrt{625}} = \frac{-15}{25} = -\frac{3}{5}$

* $\sin^2 \angle AOB + \cos^2 \angle AOB = 1$, $\sin \angle AOB = \pm\sqrt{1 - \cos^2 \angle AOB} = \pm\sqrt{1 - (-\frac{3}{5})^2} = \pm\sqrt{1 - \frac{9}{25}} = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$

* $\angle AOB$ は 0 から $\pi$ の間の角なので$\sin \angle AOB$は正の値をとる。したがって、$\sin \angle AOB = \frac{4}{5}$

6. **複素数の商と三角形の面積**

* $\alpha = 1+i, \beta = (1-\sqrt{2}) + (1+\sqrt{2})i$

* $\frac{\beta}{\alpha} = \frac{(1-\sqrt{2}) + (1+\sqrt{2})i}{1+i} = \frac{((1-\sqrt{2}) + (1+\sqrt{2})i)(1-i)}{(1+i)(1-i)} = \frac{(1-\sqrt{2}) - (1-\sqrt{2})i + (1+\sqrt{2})i - (1+\sqrt{2})i^2}{2} = \frac{1-\sqrt{2} + 1 + \sqrt{2} + i(-1+\sqrt{2}+1+\sqrt{2})}{2} = \frac{2 + 2\sqrt{2}i}{2} = 1+\sqrt{2}i$

* $O(0,0), A(1,1), B(1-\sqrt{2}, 1+\sqrt{2})$

* 三角形OABの面積 = $\frac{1}{2} |x_1y_2 - x_2y_1|$ = $\frac{1}{2} |1(1+\sqrt{2}) - 1(1-\sqrt{2})|$ = $\frac{1}{2}|1+\sqrt{2}-1+\sqrt{2}| = \frac{1}{2}|2\sqrt{2}| = \sqrt{2}$

7. **無限級数**

* $a_1 = \sqrt{3} - 1, a_2 = 4 - 2\sqrt{3}$

* 公比 $r = \frac{a_2}{a_1} = \frac{4-2\sqrt{3}}{\sqrt{3}-1} = \frac{2(2-\sqrt{3})}{\sqrt{3}-1} = \frac{2(2-\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{2(2\sqrt{3}+2-3-\sqrt{3})}{3-1} = \frac{2(\sqrt{3}-1)}{2} = \sqrt{3}-1$

* 無限級数の和 $S = \frac{a_1}{1-r} = \frac{\sqrt{3}-1}{1-(\sqrt{3}-1)} = \frac{\sqrt{3}-1}{2-\sqrt{3}} = \frac{(\sqrt{3}-1)(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{2\sqrt{3}+3-2-\sqrt{3}}{4-3} = \frac{\sqrt{3}+1}{1} = \sqrt{3}+1$

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3. 最終的な答え

1. $\frac{\sqrt{3}}{2}$

2. $\frac{\sqrt{3}}{4} + \frac{5}{4}i$

3. $a = -64, b=0$

4. $\theta = \frac{7\pi}{4}$

5. $z^5 = 41-38i$, $\sin\angle AOB = \frac{4}{5}$

6. $\frac{\beta}{\alpha} = 1+\sqrt{2}i$, 三角形OABの面積は$\sqrt{2}$

7. $\sqrt{3} + 1$