(1) $\sin \alpha + \sin \beta = \frac{1}{2}$、$\cos \alpha + \cos \beta = \frac{1}{3}$のとき、$\cos(\alpha - \beta)$の値を求める。 (2) $\tan \alpha = 2$、$\tan \beta = 4$、$\tan \gamma = 13$のとき、$\tan(\alpha + \beta + \gamma)$の値を求める。

1. 問題の内容

(1) $\sin \alpha + \sin \beta = \frac{1}{2}$、$\cos \alpha + \cos \beta = \frac{1}{3}$のとき、$\cos(\alpha - \beta)$の値を求める。

(2) $\tan \alpha = 2$、$\tan \beta = 4$、$\tan \gamma = 13$のとき、$\tan(\alpha + \beta + \gamma)$の値を求める。

2. 解き方の手順

(1)

$\sin \alpha + \sin \beta = \frac{1}{2}$ と $\cos \alpha + \cos \beta = \frac{1}{3}$ をそれぞれ2乗する。

$(\sin \alpha + \sin \beta)^2 = \sin^2 \alpha + 2\sin \alpha \sin \beta + \sin^2 \beta = \frac{1}{4}$

$(\cos \alpha + \cos \beta)^2 = \cos^2 \alpha + 2\cos \alpha \cos \beta + \cos^2 \beta = \frac{1}{9}$

2つの式を足し合わせる。

$(\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2(\sin \alpha \sin \beta + \cos \alpha \cos \beta) = \frac{1}{4} + \frac{1}{9}$

$1 + 1 + 2\cos(\alpha - \beta) = \frac{9+4}{36} = \frac{13}{36}$

$2 + 2\cos(\alpha - \beta) = \frac{13}{36}$

$2\cos(\alpha - \beta) = \frac{13}{36} - 2 = \frac{13 - 72}{36} = -\frac{59}{36}$

$\cos(\alpha - \beta) = -\frac{59}{72}$

(2)

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{2+4}{1 - 2 \cdot 4} = \frac{6}{1-8} = \frac{6}{-7} = -\frac{6}{7}$

$\tan(\alpha + \beta + \gamma) = \frac{\tan(\alpha + \beta) + \tan \gamma}{1 - \tan(\alpha + \beta) \tan \gamma} = \frac{-\frac{6}{7} + 13}{1 - (-\frac{6}{7}) \cdot 13} = \frac{-\frac{6}{7} + \frac{91}{7}}{1 + \frac{78}{7}} = \frac{\frac{85}{7}}{\frac{7+78}{7}} = \frac{85}{85} = 1$

3. 最終的な答え

(1) $\cos(\alpha - \beta) = -\frac{59}{72}$

(2) $\tan(\alpha + \beta + \gamma) = 1$