ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1024

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\[1)\cos{135{^\circ}} = \cos(90{^\circ} + 45{^\circ}) =\]

\[= \cos{90{^\circ}} \bullet \cos{45{^\circ}} -\]

\[- \sin{90{^\circ}} \bullet \sin{45{^\circ}} =\]

\[= 0 \bullet \frac{\sqrt{2}}{2} - 1 \bullet \frac{\sqrt{2}}{2} = - \frac{\sqrt{2}}{2}\]

\[2)\cos{120{^\circ}} = \cos(60{^\circ} + 60{^\circ}) =\]

\[= \cos{60{^\circ}} \bullet \cos{60{^\circ}} -\]

\[- \sin{60{^\circ}} \bullet \sin{60{^\circ}} =\]

\[= \frac{1}{2} \bullet \frac{1}{2} - \frac{\sqrt{3}}{2} \bullet \frac{\sqrt{3}}{2} = \frac{1}{4} - \frac{3}{4} =\]

\[= - \frac{2}{4} = - \frac{1}{2}\]

\[3)\cos{150{^\circ}} = \cos(90{^\circ} + 60{^\circ}) =\]

\[= \cos{90{^\circ}} \bullet \cos{60{^\circ}} -\]

\[- \sin{90{^\circ}} \bullet \sin{60{^\circ}} =\]

\[= 0 \bullet \frac{1}{2} - 1 \bullet \frac{\sqrt{3}}{2} = - \frac{\sqrt{3}}{2}\]

\[4)\cos{240{^\circ}} = \cos(180{^\circ} + 60{^\circ}) =\]

\[= \cos{180{^\circ}} \bullet \cos{60{^\circ}} -\]

\[- \sin{180{^\circ}} \bullet \sin{60{^\circ}} =\]

\[= - 1 \bullet \frac{1}{2} - 0 \bullet \frac{\sqrt{3}}{2} = - \frac{1}{2}\]