Упр.26.15 ГДЗ Мерзляк 10 класс Углубленный 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. $$\tg 75^\circ=\sqrt{\frac{1-\cos 150^\circ}{1+\cos 150^\circ}}=\sqrt{\frac{1+\frac{\sqrt3}{2}}{1-\frac{\sqrt3}{2}}}=\sqrt{\frac{2+\sqrt3}{2-\sqrt3}}=2+\sqrt3.$$

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

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

Ответ

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