\[\boxed{\mathbf{1012.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[1)\ M_{1}(0;1):\ \ \ \]
\[0^{2} + 1^{2} = 1 \Longrightarrow\]
\[\Longrightarrow M_{1} \in окружности\ (0;1).\]
\[M_{2}\left( \frac{1}{2};\frac{\sqrt{3}}{2} \right):\ \ \ \]
\[\left( \frac{1}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2} = 1\]
\[\frac{1}{4} + \frac{3}{4} = 1 \Longrightarrow\]
\[\Longrightarrow M_{2} \in окружности\ (0;1).\]
\[M_{3}\left( \frac{\sqrt{2}}{2};\frac{\sqrt{2}}{2} \right):\ \ \ \]
\[\left( \frac{\sqrt{2}}{2} \right)^{2} + \left( \frac{\sqrt{2}}{2} \right)^{2} = 1\]
\[\frac{1}{2} + \frac{1}{2} = 1 \Longrightarrow\]
\[\Longrightarrow M_{3} \in окружности\ (0;1).\]
\[M_{4}\left( - \frac{\sqrt{3}}{2};\frac{1}{2} \right):\ \left( - \frac{\sqrt{3}}{2} \right)^{2} + \left( \frac{1}{2} \right)^{2} = 1\]
\[\frac{3}{4} + \frac{1}{4} = 1 \Longrightarrow\]
\[\Longrightarrow M_{4} \in окружности\ (0;1).\]
\[A(1;0):\ \ \ \]
\[1^{2} + 0^{2} = 1 \Longrightarrow\]
\[\Longrightarrow A \in окружности\ (0;1).\]
\[B( - 1;0):\ \ \ \]
\[( - 1)^{2} + 0^{2} = 1 \Longrightarrow\]
\[\Longrightarrow \text{\ B} \in окружности\ (0;1).\]
\[2)\ Рисунок\ по\ условию\ задачи:\]
\[sin\angle AOM_{1} = sin90{^\circ} = 1;\ \]
\[cos\angle AOM_{1} = 0;\ \]
\[tg\angle AOM_{1} - не\ существует.\]
\[sin\angle AOM_{2} = \frac{\sqrt{3}}{2};\ \]
\[cos\angle AOM_{2} = \frac{1}{2};\ \]
\[tg\angle AOM_{2} = \frac{sin\angle AOM_{2}}{cos\angle AOM_{2}} = \sqrt{3}.\]
\[sin\angle AOM_{3} = \frac{\sqrt{2}}{2};\ \]
\[cos\angle AOM_{3} = \frac{\sqrt{2}}{2};\ \]
\[tg\angle AOM_{3} = \frac{sin\angle AOM_{3}}{cos\angle AOM_{3}} = 1.\]
\[sin\angle AOM_{4} = \frac{1}{2};\ \]
\[cos\angle AOM_{4} = - \frac{\sqrt{3}}{3};\ \]
\[tg\angle AOM_{4} = \frac{sin\angle AOM_{4}}{cos\angle AOM_{4}} = - \sqrt{3}.\]
\[sin\angle AOB = 0;\ \]
\[cos\angle AOB = - 1;\ \]
\[tg\angle AOB = \frac{sin\angle AOB}{cos\angle AOB} = 0.\]
\[\boxed{\mathbf{1012.еуроки - ответы\ на\ пятёрку}}\]
\[\overrightarrow{a}\ \left\{ 2;4 \right\} \Longrightarrow - \overrightarrow{a}\ \left\{ - 2;\ - 4 \right\}\]
\[\overrightarrow{b}\ \left\{ - 2;0 \right\} \Longrightarrow \ - \overrightarrow{b}\ \left\{ 2;0 \right\}\]
\[\overrightarrow{c}\ \left\{ 0;0 \right\} \Longrightarrow \ - \overrightarrow{c}\ \left\{ 0;0 \right\}\]
\[\overrightarrow{d}\ \left\{ - 2;\ - 3 \right\} \Longrightarrow - \overrightarrow{d}\ \left\{ 2;3 \right\}\]
\[\overrightarrow{e}\ \left\{ 2;\ - 3 \right\} \Longrightarrow - \overrightarrow{e}\ \left\{ - 2;3 \right\}\]
\[\overrightarrow{f}\ \left\{ 0;5 \right\} \Longrightarrow \ - \overrightarrow{f}\ \{ 0;\ - 5\}\]