Упр.25.1 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

1. Вычислите: а) $$\operatorname{tg} 15^\circ$$; б) $$\operatorname{tg} 75^\circ$$; в) $$\operatorname{tg} 105^\circ$$; г) $$\operatorname{tg} 165^\circ$$.

Используем формулу:

$$\tg(\alpha \pm \beta)=\frac{\tg \alpha \pm \tg \beta}{1 \mp \tg \alpha \cdot \tg \beta}.$$

1. $$\tg 15^\circ=\tg(45^\circ-30^\circ)=\frac{\tg 45^\circ-\tg 30^\circ}{1+\tg 45^\circ\cdot \tg 30^\circ}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$$

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

2. $$\tg 75^\circ=\tg(45^\circ+30^\circ)=\frac{\tg 45^\circ+\tg 30^\circ}{1-\tg 45^\circ\cdot \tg 30^\circ}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$$

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

3. $$\tg 105^\circ=\tg(45^\circ+60^\circ)=\frac{\tg 45^\circ+\tg 60^\circ}{1-\tg 45^\circ\cdot \tg 60^\circ}=\frac{1+\sqrt{3}}{1-\sqrt{3}}$$

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

4. $$\tg 165^\circ=\tg(45^\circ+120^\circ)=\frac{\tg 45^\circ+\tg 120^\circ}{1-\tg 45^\circ\cdot \tg 120^\circ}.$$

   Так как $$\tg 120^\circ=\tg(90^\circ+30^\circ)=-\ctg 30^\circ=-\sqrt{3},$$

   то

   $$\tg 165^\circ=\frac{1-\sqrt{3}}{1+\sqrt{3}}=\frac{(1-\sqrt{3})^2}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{1-2\sqrt{3}+3}{1-3}=\sqrt{3}-2.$$

Ответ: $$\tg 15^\circ=2-\sqrt{3},\ \tg 75^\circ=2+\sqrt{3},\ \tg 105^\circ=-2-\sqrt{3},\ \tg 165^\circ=\sqrt{3}-2.$$