与えられた3つのcos関数の式をsin関数に変形する問題です。 (1) $\cos(x + \frac{\pi}{6})$ (2) $\cos(x - \frac{\pi}{3})$ (3) $\cos(2x + \frac{\pi}{6})$

1. 問題の内容

与えられた3つのcos関数の式をsin関数に変形する問題です。

(1) $\cos(x + \frac{\pi}{6})$

(2) $\cos(x - \frac{\pi}{3})$

(3) $\cos(2x + \frac{\pi}{6})$

2. 解き方の手順

cos関数をsin関数に変換するには、以下の公式を利用します。

$\cos(\theta) = \sin(\frac{\pi}{2} - \theta)$

(1) $\cos(x + \frac{\pi}{6})$

$\cos(x + \frac{\pi}{6}) = \sin(\frac{\pi}{2} - (x + \frac{\pi}{6})) = \sin(\frac{\pi}{2} - x - \frac{\pi}{6}) = \sin(\frac{3\pi}{6} - \frac{\pi}{6} - x) = \sin(\frac{2\pi}{6} - x) = \sin(\frac{\pi}{3} - x)$

(2) $\cos(x - \frac{\pi}{3})$

$\cos(x - \frac{\pi}{3}) = \sin(\frac{\pi}{2} - (x - \frac{\pi}{3})) = \sin(\frac{\pi}{2} - x + \frac{\pi}{3}) = \sin(\frac{3\pi}{6} + \frac{2\pi}{6} - x) = \sin(\frac{5\pi}{6} - x)$

(3) $\cos(2x + \frac{\pi}{6})$

$\cos(2x + \frac{\pi}{6}) = \sin(\frac{\pi}{2} - (2x + \frac{\pi}{6})) = \sin(\frac{\pi}{2} - 2x - \frac{\pi}{6}) = \sin(\frac{3\pi}{6} - \frac{\pi}{6} - 2x) = \sin(\frac{2\pi}{6} - 2x) = \sin(\frac{\pi}{3} - 2x)$

3. 最終的な答え

(1) $\sin(\frac{\pi}{3} - x)$

(2) $\sin(\frac{5\pi}{6} - x)$

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