Выполни упрощение тригонометрических выражений №1 (1-7), а также найди синус и тангенс, если косинус равен sqrt(2)/2, косинус и синус, если тангенс равен 8, косинус и тангенс, если синус равен 1/2.

Задание №1 1) $ctg(u)(1-cos^2(u)) = ctg(u)sin^2(u) = \frac{cos(u)}{sin(u)}sin^2(u) = cos(u)sin(u)$ 2) $1 - sin^2(\alpha) - cos^2(u) = cos^2(\alpha) - cos^2(u)$ 3) $sin(\beta)ctg(\beta) - cos(\beta) = sin(\beta)\frac{cos(\beta)}{sin(\beta)} - cos(\beta) = cos(\beta) - cos(\beta) = 0$ 4) $(1 - sin(\beta))(1 + sin(\beta)) + (1 + cos(\beta))(1 - cos(\beta)) = 1 - sin^2(\beta) + 1 - cos^2(\beta) = cos^2(\beta) + sin^2(\beta) = 1 + 1 - 1 = 1$ 5) $\frac{cos(\alpha)}{1 + sin(\alpha)} + \frac{cos(\alpha)}{1 - sin(\alpha)} = \frac{cos(\alpha)(1 - sin(\alpha)) + cos(\alpha)(1 + sin(\alpha))}{(1 + sin(\alpha))(1 - sin(\alpha))} = \frac{cos(\alpha) - cos(\alpha)sin(\alpha) + cos(\alpha) + cos(\alpha)sin(\alpha))}{1 - sin^2(\alpha)} = \frac{2cos(\alpha)}{cos^2(\alpha)} = \frac{2}{cos(\alpha)}$ 6) $\frac{cos(u)}{tg(u) + \frac{cos(u)}{1 + sin(u)}} = \frac{cos(u)}{\frac{sin(u)}{cos(u)} + \frac{cos(u)}{1 + sin(u)}} = \frac{cos(u)}{\frac{sin(u)(1 + sin(u)) + cos^2(u)}{cos(u)(1 + sin(u))}} = \frac{cos(u)(cos(u)(1 + sin(u)))}{sin(u) + sin^2(u) + cos^2(u)} = \frac{cos^2(u)(1 + sin(u))}{sin(u) + 1} = cos^2(u)$ 7) $\frac{1}{1 + ctg(u)} - \frac{1}{1 - ctg(u)} = \frac{1 - ctg(u) - (1 + ctg(u))}{(1 + ctg(u))(1 - ctg(u))} = \frac{-2ctg(u)}{1 - ctg^2(u)}$ Задание №2 $cos(\alpha) = \frac{\sqrt{2}}{2}$ $sin^2(\alpha) + cos^2(\alpha) = 1$ $sin^2(\alpha) = 1 - cos^2(\alpha) = 1 - (\frac{\sqrt{2}}{2})^2 = 1 - \frac{2}{4} = \frac{1}{2}$ $sin(\alpha) = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}$ $tg(\alpha) = \frac{sin(\alpha)}{cos(\alpha)} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1$ Задание №3 **Допущение:** $tg \alpha = 3$ $tg(\alpha) = \frac{sin(\alpha)}{cos(\alpha)}$ $\frac{sin(\alpha)}{cos(\alpha)} = 3$ $sin(\alpha) = 3cos(\alpha)$ $sin^2(\alpha) + cos^2(\alpha) = 1$ $(3cos(\alpha))^2 + cos^2(\alpha) = 1$ $9cos^2(\alpha) + cos^2(\alpha) = 1$ $10cos^2(\alpha) = 1$ $cos^2(\alpha) = \frac{1}{10}$ $cos(\alpha) = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$ $sin(\alpha) = 3cos(\alpha) = 3 * \frac{\sqrt{10}}{10} = \frac{3\sqrt{10}}{10}$ Задание №4 $sin(\alpha) = \frac{1}{2}$ $sin^2(\alpha) + cos^2(\alpha) = 1$ $cos^2(\alpha) = 1 - sin^2(\alpha) = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$ $cos(\alpha) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ $tg(\alpha) = \frac{sin(\alpha)}{cos(\alpha)} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$