与えられた三角関数の式を $r\sin(\theta + \alpha)$ の形に変形し、$r$ と $\alpha$ の値を求める問題です。ただし、$r>0$、$ -\pi \leqq \alpha \leqq \pi$ とします。$\alpha$ の値を求めることが困難な場合は、$\alpha$が満たすべき条件を記します。

はい、承知しました。画像にある三角関数の合成の問題を解きます。

1. 問題の内容

与えられた三角関数の式を $r\sin(\theta + \alpha)$ の形に変形し、$r$ と $\alpha$ の値を求める問題です。ただし、$r>0$、$-\pi \leqq \alpha \leqq \pi$ とします。$\alpha$ の値を求めることが困難な場合は、$\alpha$が満たすべき条件を記します。

2. 解き方の手順

三角関数の合成の公式を利用します。一般に、$a\sin\theta + b\cos\theta = r\sin(\theta + \alpha)$ と表せます。ここで、$r = \sqrt{a^2 + b^2}$ であり、$\cos\alpha = \frac{a}{r}$、$\sin\alpha = \frac{b}{r}$ となります。

(1) $\sin\theta + \cos\theta$

$a = 1$, $b = 1$ より、$r = \sqrt{1^2 + 1^2} = \sqrt{2}$

$\cos\alpha = \frac{1}{\sqrt{2}}$, $\sin\alpha = \frac{1}{\sqrt{2}}$ なので、$\alpha = \frac{\pi}{4}$

答え: $\sqrt{2}\sin(\theta + \frac{\pi}{4})$

(2) $\sin\theta - \cos\theta$

$a = 1$, $b = -1$ より、$r = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

$\cos\alpha = \frac{1}{\sqrt{2}}$, $\sin\alpha = \frac{-1}{\sqrt{2}}$ なので、$\alpha = -\frac{\pi}{4}$

答え: $\sqrt{2}\sin(\theta - \frac{\pi}{4})$

(3) $\sqrt{3}\sin\theta - \cos\theta$

$a = \sqrt{3}$, $b = -1$ より、$r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2$

$\cos\alpha = \frac{\sqrt{3}}{2}$, $\sin\alpha = \frac{-1}{2}$ なので、$\alpha = -\frac{\pi}{6}$

答え: $2\sin(\theta - \frac{\pi}{6})$

(4) $-\sqrt{3}\sin\theta + \cos\theta$

$a = -\sqrt{3}$, $b = 1$ より、$r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$

$\cos\alpha = \frac{-\sqrt{3}}{2}$, $\sin\alpha = \frac{1}{2}$ なので、$\alpha = \frac{5\pi}{6}$

答え: $2\sin(\theta + \frac{5\pi}{6})$

(5) $\sin\theta + \sqrt{3}\cos\theta$

$a = 1$, $b = \sqrt{3}$ より、$r = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$

$\cos\alpha = \frac{1}{2}$, $\sin\alpha = \frac{\sqrt{3}}{2}$ なので、$\alpha = \frac{\pi}{3}$

答え: $2\sin(\theta + \frac{\pi}{3})$

(6) $-\sin\theta + \sqrt{3}\cos\theta$

$a = -1$, $b = \sqrt{3}$ より、$r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$

$\cos\alpha = \frac{-1}{2}$, $\sin\alpha = \frac{\sqrt{3}}{2}$ なので、$\alpha = \frac{2\pi}{3}$

答え: $2\sin(\theta + \frac{2\pi}{3})$

(7) $2\sin\theta + 2\cos\theta$

$a = 2$, $b = 2$ より、$r = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$

$\cos\alpha = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$, $\sin\alpha = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$ なので、$\alpha = \frac{\pi}{4}$

答え: $2\sqrt{2}\sin(\theta + \frac{\pi}{4})$

(8) $\sqrt{6}\sin\theta + \sqrt{2}\cos\theta$

$a = \sqrt{6}$, $b = \sqrt{2}$ より、$r = \sqrt{(\sqrt{6})^2 + (\sqrt{2})^2} = \sqrt{6+2} = \sqrt{8} = 2\sqrt{2}$

$\cos\alpha = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2}$, $\sin\alpha = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}$ なので、$\alpha = \frac{\pi}{6}$

答え: $2\sqrt{2}\sin(\theta + \frac{\pi}{6})$

(9) $\sqrt{3}\sin\theta - 3\cos\theta$

$a = \sqrt{3}$, $b = -3$ より、$r = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3+9} = \sqrt{12} = 2\sqrt{3}$

$\cos\alpha = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$, $\sin\alpha = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$ なので、$\alpha = -\frac{\pi}{3}$

答え: $2\sqrt{3}\sin(\theta - \frac{\pi}{3})$

(10) $2\sin\theta - 3\cos\theta$

$a = 2$, $b = -3$ より、$r = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$

$\cos\alpha = \frac{2}{\sqrt{13}}$, $\sin\alpha = \frac{-3}{\sqrt{13}}$ なので、$\alpha = \arctan(-\frac{3}{2})$ となり、具体的な角度を求めるのは難しい。$-\frac{\pi}{2} < \alpha < 0$

答え: $\sqrt{13}\sin(\theta + \alpha)$, where $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = -\frac{3}{\sqrt{13}}$, or $-\frac{\pi}{2} < \alpha < 0$

(11) $5\sin\theta + 12\cos\theta$

$a = 5$, $b = 12$ より、$r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

$\cos\alpha = \frac{5}{13}$, $\sin\alpha = \frac{12}{13}$ なので、$\alpha = \arctan(\frac{12}{5})$ となり、具体的な角度を求めるのは難しい。$0 < \alpha < \frac{\pi}{2}$

答え: $13\sin(\theta + \alpha)$, where $\cos \alpha = \frac{5}{13}$ and $\sin \alpha = \frac{12}{13}$, or $0 < \alpha < \frac{\pi}{2}$

(12) $\sqrt{2}\sin\theta + \cos\theta$

$a = \sqrt{2}$, $b = 1$ より、$r = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3}$

$\cos\alpha = \frac{\sqrt{2}}{\sqrt{3}}$, $\sin\alpha = \frac{1}{\sqrt{3}}$ なので、$\alpha = \arctan(\frac{1}{\sqrt{2}})$ となり、具体的な角度を求めるのは難しい。$0 < \alpha < \frac{\pi}{2}$

答え: $\sqrt{3}\sin(\theta + \alpha)$, where $\cos \alpha = \frac{\sqrt{2}}{\sqrt{3}}$ and $\sin \alpha = \frac{1}{\sqrt{3}}$, or $0 < \alpha < \frac{\pi}{2}$

(13) $\frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}\cos\theta$

$a = \frac{1}{2}$, $b = -\frac{\sqrt{3}}{2}$ より、$r = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

$\cos\alpha = \frac{1}{2}$, $\sin\alpha = -\frac{\sqrt{3}}{2}$ なので、$\alpha = -\frac{\pi}{3}$

答え: $\sin(\theta - \frac{\pi}{3})$

(14) $\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta$

$a = \frac{1}{\sqrt{2}}$, $b = \frac{1}{\sqrt{2}}$ より、$r = \sqrt{(\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1$

$\cos\alpha = \frac{1}{\sqrt{2}}$, $\sin\alpha = \frac{1}{\sqrt{2}}$ なので、$\alpha = \frac{\pi}{4}$

答え: $\sin(\theta + \frac{\pi}{4})$

3. 最終的な答え

(1) $\sqrt{2}\sin(\theta + \frac{\pi}{4})$

(2) $\sqrt{2}\sin(\theta - \frac{\pi}{4})$

(3) $2\sin(\theta - \frac{\pi}{6})$

(4) $2\sin(\theta + \frac{5\pi}{6})$

(5) $2\sin(\theta + \frac{\pi}{3})$

(6) $2\sin(\theta + \frac{2\pi}{3})$

(7) $2\sqrt{2}\sin(\theta + \frac{\pi}{4})$

(8) $2\sqrt{2}\sin(\theta + \frac{\pi}{6})$

(9) $2\sqrt{3}\sin(\theta - \frac{\pi}{3})$

(10) $\sqrt{13}\sin(\theta + \alpha)$, where $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = -\frac{3}{\sqrt{13}}$, or $-\frac{\pi}{2} < \alpha < 0$

(11) $13\sin(\theta + \alpha)$, where $\cos \alpha = \frac{5}{13}$ and $\sin \alpha = \frac{12}{13}$, or $0 < \alpha < \frac{\pi}{2}$

(12) $\sqrt{3}\sin(\theta + \alpha)$, where $\cos \alpha = \frac{\sqrt{2}}{\sqrt{3}}$ and $\sin \alpha = \frac{1}{\sqrt{3}}$, or $0 < \alpha < \frac{\pi}{2}$

(13) $\sin(\theta - \frac{\pi}{3})$

(14) $\sin(\theta + \frac{\pi}{4})$