Упр.8.57 ГДЗ Погорелов 7-9 класс (Геометрия)

Найдите $$\cos\alpha$$ и $$\operatorname{ctg}\alpha$$, если:

1. $$\sin\alpha=0{,}6$$, $$0^\circ<\alpha<90^\circ$$;
2. $$\sin\alpha=\frac{1}{\sqrt{2}}$$, $$90^\circ<\alpha<180^\circ$$;
3. $$\sin\alpha=0^\circ<\alpha<180^\circ$$.

Используем тождества:

$$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}, \qquad \sin^2 \alpha+\cos^2 \alpha=1,$$

откуда

$$\cos \alpha=\pm\sqrt{1-\sin^2 \alpha}.$$

1. Если $$\sin \alpha=0{,}6=\frac{3}{5}, \quad 0^\circ<\alpha<90^\circ,$$ то $$\cos \alpha>0.$$

   $$\cos \alpha=\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}.$$

   Тогда

   $$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}.$$

2. Если $$\sin \alpha=\frac{1}{3}, \quad 90^\circ<\alpha<180^\circ,$$ то $$\cos \alpha<0.$$

   $$\cos \alpha=-\sqrt{1-\left(\frac{1}{3}\right)^2}=-\sqrt{1-\frac{1}{9}}=-\sqrt{\frac{8}{9}}=-\frac{2\sqrt{2}}{3}.$$

   Тогда

   $$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}}=-2\sqrt{2}.$$

3. Если $$\sin \alpha=\frac{1}{\sqrt{2}}, \quad 0^\circ<\alpha<180^\circ,$$ то

   $$\cos \alpha=\pm\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^2}=\pm\sqrt{1-\frac{1}{2}}=\pm\sqrt{\frac{1}{2}}=\pm\frac{1}{\sqrt{2}}.$$

   Следовательно,

   $$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{\pm\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=\pm 1.$$

Ответ: 1) $$\cos \alpha=\frac{4}{5}, \ \ctg \alpha=\frac{4}{3};$$ 2) $$\cos \alpha=-\frac{2\sqrt{2}}{3}, \ \ctg \alpha=-2\sqrt{2};$$ 3) $$\cos \alpha=\pm\frac{1}{\sqrt{2}}, \ \ctg \alpha=\pm 1.$$