$\tan a = 3$、$\sin \beta = \frac{1}{4}$のとき、$\cos(a+\beta)$, $\sin(a+\beta)$, $\tan(a+\beta)$の値を求めよ。

1. 問題の内容

$\tan a = 3$、$\sin \beta = \frac{1}{4}$のとき、$\cos(a+\beta)$, $\sin(a+\beta)$, $\tan(a+\beta)$の値を求めよ。

2. 解き方の手順

まず、$\tan a = 3$から$\cos a$と$\sin a$の値を求める。$1 + \tan^2 a = \frac{1}{\cos^2 a}$より、$\cos^2 a = \frac{1}{1 + \tan^2 a} = \frac{1}{1+3^2} = \frac{1}{10}$となる。したがって、$\cos a = \pm \frac{1}{\sqrt{10}}$。

$\sin a = \tan a \cos a = 3 \cos a = \pm \frac{3}{\sqrt{10}}$.

次に、$\sin \beta = \frac{1}{4}$から$\cos \beta$の値を求める。$\sin^2 \beta + \cos^2 \beta = 1$より、$\cos^2 \beta = 1 - \sin^2 \beta = 1 - (\frac{1}{4})^2 = 1 - \frac{1}{16} = \frac{15}{16}$。したがって、$\cos \beta = \pm \frac{\sqrt{15}}{4}$.

$\cos(a+\beta) = \cos a \cos \beta - \sin a \sin \beta = (\pm \frac{1}{\sqrt{10}})(\pm \frac{\sqrt{15}}{4}) - (\pm \frac{3}{\sqrt{10}})(\frac{1}{4}) = \pm \frac{\sqrt{15}}{4\sqrt{10}} \mp \frac{3}{4\sqrt{10}} = \frac{\pm \sqrt{15} \mp 3}{4\sqrt{10}} = \frac{\pm \sqrt{6} \mp \frac{3}{\sqrt{5}}}{4\sqrt{2}}$

$\sin(a+\beta) = \sin a \cos \beta + \cos a \sin \beta = (\pm \frac{3}{\sqrt{10}})(\pm \frac{\sqrt{15}}{4}) + (\pm \frac{1}{\sqrt{10}})(\frac{1}{4}) = \pm \frac{3\sqrt{15}}{4\sqrt{10}} \pm \frac{1}{4\sqrt{10}} = \frac{\pm 3\sqrt{15} \pm 1}{4\sqrt{10}}$

$\tan(a+\beta) = \frac{\tan a + \tan \beta}{1 - \tan a \tan \beta}$.

$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{1}{4}}{\pm \frac{\sqrt{15}}{4}} = \pm \frac{1}{\sqrt{15}} = \pm \frac{\sqrt{15}}{15}$.

したがって、$\tan(a+\beta) = \frac{3 \pm \frac{\sqrt{15}}{15}}{1 - 3(\pm \frac{\sqrt{15}}{15})} = \frac{3 \pm \frac{\sqrt{15}}{15}}{1 \mp \frac{\sqrt{15}}{5}} = \frac{45 \pm \sqrt{15}}{15} \cdot \frac{5}{5 \mp \sqrt{15}} = \frac{45 \pm \sqrt{15}}{3(5 \mp \sqrt{15})} = \frac{(45 \pm \sqrt{15})(5 \pm \sqrt{15})}{3(25-15)} = \frac{225 \pm 45\sqrt{15} + 5\sqrt{15} \pm 15}{30} = \frac{240 \pm 50\sqrt{15}}{30} = \frac{24 \pm 5\sqrt{15}}{3}$.

特に条件がないので、$\cos a, \cos \beta$の正負の組み合わせで複数の答えが出る。ここでは、$a$と$\beta$がともに鋭角であると仮定する。つまり、$\cos a = \frac{1}{\sqrt{10}}$, $\cos \beta = \frac{\sqrt{15}}{4}$.

$\cos(a+\beta) = \frac{\sqrt{15}}{4\sqrt{10}} - \frac{3}{4\sqrt{10}} = \frac{\sqrt{15}-3}{4\sqrt{10}} = \frac{\sqrt{6}-\frac{3}{\sqrt{5}}}{4\sqrt{2}}$

$\sin(a+\beta) = \frac{3\sqrt{15}}{4\sqrt{10}} + \frac{1}{4\sqrt{10}} = \frac{3\sqrt{15}+1}{4\sqrt{10}}$

$\tan(a+\beta) = \frac{24-5\sqrt{15}}{3}$

3. 最終的な答え

$\cos(a+\beta) = \frac{\sqrt{15}-3}{4\sqrt{10}}$

$\sin(a+\beta) = \frac{3\sqrt{15}+1}{4\sqrt{10}}$

$\tan(a+\beta) = \frac{24-5\sqrt{15}}{3}$