三角関数の変形に関する問題です。 前半は$\cos$関数を$\sin$関数に変形し、後半は$\sin$関数を$\cos$関数に変形します。 公式 $\cos x = \sin(x + \frac{\pi}{2})$ と $\sin x = \cos(x - \frac{\pi}{2})$ を利用します。

1. 問題の内容

三角関数の変形に関する問題です。

前半は$\cos$関数を$\sin$関数に変形し、後半は$\sin$関数を$\cos$関数に変形します。

公式 $\cos x = \sin(x + \frac{\pi}{2})$ と $\sin x = \cos(x - \frac{\pi}{2})$ を利用します。

2. 解き方の手順

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

$\cos x = \sin(x + \frac{\pi}{2})$を利用すると、

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

$\sin(x + \frac{\pi}{6} + \frac{3\pi}{6}) = \sin(x + \frac{4\pi}{6}) = \sin(x + \frac{2\pi}{3})$

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

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

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

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

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

$\sin(2x + \frac{\pi}{6} + \frac{3\pi}{6}) = \sin(2x + \frac{4\pi}{6}) = \sin(2x + \frac{2\pi}{3})$

(4) $\sin(x - \frac{3\pi}{4})$

$\sin x = \cos(x - \frac{\pi}{2})$を利用すると、

$\sin(x - \frac{3\pi}{4}) = \cos(x - \frac{3\pi}{4} - \frac{\pi}{2})$

$\cos(x - \frac{3\pi}{4} - \frac{2\pi}{4}) = \cos(x - \frac{5\pi}{4})$

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

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

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

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

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

$\cos(x - \frac{\pi}{6} - \frac{3\pi}{6}) = \cos(x - \frac{4\pi}{6}) = \cos(x - \frac{2\pi}{3})$

3. 最終的な答え

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

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

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

(4) $\cos(x - \frac{5\pi}{4})$

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

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