与えられた三角関数の式を簡略化します。具体的には、$\sin(x + \pi)$, $\cos(x + \pi)$, $\tan(x + \pi)$, $\sin(x + \frac{\pi}{2})$, $\cos(x + \frac{\pi}{2})$, $\tan(x + \frac{\pi}{2})$ をそれぞれ計算します。

1. 問題の内容

与えられた三角関数の式を簡略化します。具体的には、$\sin(x + \pi)$, $\cos(x + \pi)$, $\tan(x + \pi)$, $\sin(x + \frac{\pi}{2})$, $\cos(x + \frac{\pi}{2})$, $\tan(x + \frac{\pi}{2})$ をそれぞれ計算します。

2. 解き方の手順

加法定理および三角関数の性質を利用します。

* $\sin(x + \pi)$ について：

$\sin(x + \pi) = \sin(x)\cos(\pi) + \cos(x)\sin(\pi) = \sin(x)(-1) + \cos(x)(0) = -\sin(x)$

* $\cos(x + \pi)$ について：

$\cos(x + \pi) = \cos(x)\cos(\pi) - \sin(x)\sin(\pi) = \cos(x)(-1) - \sin(x)(0) = -\cos(x)$

* $\tan(x + \pi)$ について：

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

* $\sin(x + \frac{\pi}{2})$ について：

$\sin(x + \frac{\pi}{2}) = \sin(x)\cos(\frac{\pi}{2}) + \cos(x)\sin(\frac{\pi}{2}) = \sin(x)(0) + \cos(x)(1) = \cos(x)$

* $\cos(x + \frac{\pi}{2})$ について：

$\cos(x + \frac{\pi}{2}) = \cos(x)\cos(\frac{\pi}{2}) - \sin(x)\sin(\frac{\pi}{2}) = \cos(x)(0) - \sin(x)(1) = -\sin(x)$

* $\tan(x + \frac{\pi}{2})$ について：

$\tan(x + \frac{\pi}{2}) = \frac{\sin(x + \frac{\pi}{2})}{\cos(x + \frac{\pi}{2})} = \frac{\cos(x)}{-\sin(x)} = -\cot(x) = -\frac{1}{\tan(x)}$

3. 最終的な答え

$\sin(x + \pi) = -\sin(x)$

$\cos(x + \pi) = -\cos(x)$

$\tan(x + \pi) = \tan(x)$

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

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

$\tan(x + \frac{\pi}{2}) = -\frac{1}{\tan(x)}$