Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 525

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\[1)\cos{150{^\circ}} = \cos(90{^\circ} + 60{^\circ}) =\]

\[= - \sin{60{^\circ}} = - \frac{\sqrt{3}}{2}\]

\[2)\sin{135{^\circ}} = \sin(180{^\circ} - 135{^\circ}) =\]

\[= \sin{45{^\circ}} = \frac{\sqrt{2}}{2}\]

\[3)\ ctg\ 135{^\circ} = ctg(90{^\circ} + 45{^\circ}) =\]

\[= - tg\ 45{^\circ} = - 1\]

\[4)\cos{120{^\circ}} = \cos(90{^\circ} + 30{^\circ}) =\]

\[= - \sin{30{^\circ}} = - \frac{1}{2}\]

\[5)\cos{225{^\circ}} = \cos(180{^\circ} + 45{^\circ}) =\]

\[= - \cos{45{^\circ}} = - \frac{\sqrt{2}}{2}\]

\[6)\sin{210{^\circ}} = \sin(180{^\circ} + 30{^\circ}) =\]

\[= - \sin{30{^\circ}} = - \frac{1}{2}\]

\[7)\ ctg\ 240{^\circ} =\]

\[= \text{ctg\ }(180{^\circ} + 60{^\circ}) = ctg\ 60{^\circ} =\]

\[= \frac{\sqrt{3}}{3}\]

\[8)\sin{315{^\circ}} = \sin(270{^\circ} + 45{^\circ}) =\]

\[= - \cos{45{^\circ}} = - \frac{\sqrt{2}}{2}\]