Реши задачу: Углы а и в смежные, cos a = -1/6. Найди cos ẞ.

3. Углы \(\alpha\) и \(\beta\) смежные, \(\cos \alpha = -\frac{1}{6}\) 1) Найдите \(\cos \beta\). \(\cos \beta = -\cos \alpha = \frac{1}{6}\) 2) Какой из углов \(\alpha\) и \(\beta\) является острым, а какой тупым? Т.к. \(\cos \alpha < 0\), то угол \(\alpha\) тупой, а угол \(\beta\) острый. 4. Найдите значение выражения: 1) \(2\sin 90^\circ + 3\cos 0^\circ = 2 \cdot 1 + 3 \cdot 1 = 2 + 3 = 5\) 2) \(3\sin 0^\circ - 5\cos 180^\circ = 3 \cdot 0 - 5 \cdot (-1) = 0 + 5 = 5\) 3) \(\operatorname{tg} 23^\circ \cdot \operatorname{tg} 0^\circ \cdot \operatorname{tg} 106^\circ = \operatorname{tg} 23^\circ \cdot 0 \cdot \operatorname{tg} 106^\circ = 0\) 4) \(6 \operatorname{tg} 180^\circ + 5 = 6 \cdot 0 + 5 = 0 + 5 = 5\) 5) \(\cos^2 165^\circ + \sin^2 165^\circ = 1\) (основное тригонометрическое тождество) 6) \(\frac{\sin 0^\circ + \sin 90^\circ}{\cos 0^\circ - \cos 90^\circ} = \frac{0 + 1}{1 - 0} = \frac{1}{1} = 1\) 5. Вычислите: 1) \(4\cos 90^\circ + 2\cos 180^\circ - \operatorname{ctg} 90^\circ = 4 \cdot 0 + 2 \cdot (-1) - 0 = 0 - 2 - 0 = -2\) *Перевод:* 3. Angles α and β are adjacent, cos α = -1/6 1) Find cos β. cos β = -cos α = 1/6 2) Which of the angles α and β is acute and which is obtuse? Since cos α < 0, the angle α is obtuse and the angle β is acute. 4. Find the value of the expression: 1) 2sin 90° + 3cos 0° = 2 ⋅ 1 + 3 ⋅ 1 = 2 + 3 = 5 2) 3sin 0° - 5cos 180° = 3 ⋅ 0 - 5 ⋅ (-1) = 0 + 5 = 5 3) tg 23° ⋅ tg 0° ⋅ tg 106° = tg 23° ⋅ 0 ⋅ tg 106° = 0 4) 6 tg 180° + 5 = 6 ⋅ 0 + 5 = 0 + 5 = 5 5) cos² 165° + sin² 165° = 1 (basic trigonometric identity) 6) (sin 0° + sin 90°) / (cos 0° - cos 90°) = (0 + 1) / (1 - 0) = 1/1 = 1 7. Calculate: 1) 4cos 90° + 2cos 180° - ctg 90° = 4 ⋅ 0 + 2 ⋅ (-1) - 0 = 0 - 2 - 0 = -2