次の3つの式を簡単にせよ。 (1) $(1 + \sin 45^\circ + \sin 150^\circ)(1 + \cos 135^\circ + \cos 60^\circ)$ (2) $\cos 150^\circ \tan 30^\circ - \tan 135^\circ \sin 30^\circ$ (3) $\sin(90^\circ - \theta) \cos(180^\circ - \theta) - \cos^2 \theta \tan^2(180^\circ - \theta)$

はい、承知いたしました。数学の問題を解いていきます。

1. 問題の内容

次の3つの式を簡単にせよ。

(1) $(1 + \sin 45^\circ + \sin 150^\circ)(1 + \cos 135^\circ + \cos 60^\circ)$

(2) $\cos 150^\circ \tan 30^\circ - \tan 135^\circ \sin 30^\circ$

(3) $\sin(90^\circ - \theta) \cos(180^\circ - \theta) - \cos^2 \theta \tan^2(180^\circ - \theta)$

2. 解き方の手順

(1)

$\sin 45^\circ = \frac{\sqrt{2}}{2}$

$\sin 150^\circ = \sin (180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$

$\cos 135^\circ = \cos (180^\circ - 45^\circ) = - \cos 45^\circ = -\frac{\sqrt{2}}{2}$

$\cos 60^\circ = \frac{1}{2}$

$(1 + \frac{\sqrt{2}}{2} + \frac{1}{2})(1 - \frac{\sqrt{2}}{2} + \frac{1}{2}) = (\frac{3}{2} + \frac{\sqrt{2}}{2})(\frac{3}{2} - \frac{\sqrt{2}}{2}) = (\frac{3}{2})^2 - (\frac{\sqrt{2}}{2})^2 = \frac{9}{4} - \frac{2}{4} = \frac{7}{4}$

(2)

$\cos 150^\circ = \cos(180^\circ - 30^\circ) = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$

$\tan 30^\circ = \frac{1}{\sqrt{3}}$

$\tan 135^\circ = \tan(180^\circ - 45^\circ) = -\tan 45^\circ = -1$

$\sin 30^\circ = \frac{1}{2}$

$(-\frac{\sqrt{3}}{2})(\frac{1}{\sqrt{3}}) - (-1)(\frac{1}{2}) = -\frac{1}{2} + \frac{1}{2} = 0$

(3)

$\sin(90^\circ - \theta) = \cos \theta$

$\cos(180^\circ - \theta) = - \cos \theta$

$\tan(180^\circ - \theta) = - \tan \theta$

$\cos \theta (-\cos \theta) - \cos^2 \theta (-\tan \theta)^2 = - \cos^2 \theta - \cos^2 \theta \tan^2 \theta = -\cos^2 \theta - \cos^2 \theta \frac{\sin^2 \theta}{\cos^2 \theta} = -\cos^2 \theta - \sin^2 \theta = -(\cos^2 \theta + \sin^2 \theta) = -1$

3. 最終的な答え

(1) $\frac{7}{4}$

(2) $0$

(3) $-1$