Упр.23.21 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) sin 15°; 3) tg 75°; 5) tg 112°30′;
2) cos 15°; 4) cos 75°; 6) tg ?/8.

1. $$\sin 15^\circ=\sqrt{\frac{1-\cos 30^\circ}{2}}=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\sqrt{\frac{2-\sqrt3}{4}}=\frac{\sqrt{2-\sqrt3}}{2}.$$

2. $$\cos 15^\circ=\sqrt{\frac{1+\cos 30^\circ}{2}}=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}=\sqrt{\frac{2+\sqrt3}{4}}=\frac{\sqrt{2+\sqrt3}}{2}.$$

3. $$\tg 75^\circ=\sqrt{\frac{1-\cos 150^\circ}{1+\cos 150^\circ}}=\sqrt{\frac{1-(-\frac{\sqrt3}{2})}{1+(-\frac{\sqrt3}{2})}}=\sqrt{\frac{1+\frac{\sqrt3}{2}}{1-\frac{\sqrt3}{2}}}.$$

   $$\tg 75^\circ=\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}=\sqrt{\frac{(2+\sqrt3)^2}{4-3}}=2+\sqrt3.$$

4. $$\cos 75^\circ=\sqrt{\frac{1+\cos 150^\circ}{2}}=\sqrt{\frac{1-\frac{\sqrt3}{2}}{2}}=\sqrt{\frac{2-\sqrt3}{4}}=\frac{\sqrt{2-\sqrt3}}{2}.$$

5. $$\tg 112^\circ 30’=-\sqrt{\frac{1-\cos 225^\circ}{1+\cos 225^\circ}}=-\sqrt{\frac{1-(-\frac{\sqrt2}{2})}{1+(-\frac{\sqrt2}{2})}}=-\sqrt{\frac{1+\frac{\sqrt2}{2}}{1-\frac{\sqrt2}{2}}}.$$

   $$\tg 112^\circ 30’=-\sqrt{\frac{2+\sqrt2}{2-\sqrt2}}=-\sqrt{\frac{(\sqrt2+1)^2}{1}}=-(\sqrt2+1).$$

6. $$\tg \frac{\pi}{8}=\sqrt{\frac{1-\cos \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}}=\sqrt{\frac{1-\frac{\sqrt2}{2}}{1+\frac{\sqrt2}{2}}}=\sqrt{\frac{2-\sqrt2}{2+\sqrt2}}.$$

   $$\tg \frac{\pi}{8}=\sqrt{\frac{(\sqrt2-1)^2}{1}}=\sqrt2-1.$$

Ответ

$$\sin 15^\circ=\frac{\sqrt{2-\sqrt3}}{2},\quad \cos 15^\circ=\frac{\sqrt{2+\sqrt3}}{2},\quad \tg 75^\circ=2+\sqrt3,$$
$$\cos 75^\circ=\frac{\sqrt{2-\sqrt3}}{2},\quad \tg 112^\circ 30’=-(\sqrt2+1),\quad \tg \frac{\pi}{8}=\sqrt2-1.$$