ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 529

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\[1)\cos{750{^\circ}} =\]

\[= \cos(2 \bullet 360{^\circ} + 30{^\circ}) =\]

\[= \cos{30{^\circ}} = \frac{\sqrt{3}}{2}\]

\[2)\sin{1140{^\circ}} =\]

\[= \sin(3 \bullet 360{^\circ} + 60{^\circ}) =\]

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

\[3)\ tg\ 405{^\circ} =\]

\[= \text{tg}(2 \bullet 180{^\circ} + 45{^\circ}) =\]

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

\[4)\cos{840{^\circ}} =\]

\[= \cos(2 \bullet 360{^\circ} + 120{^\circ}) =\]

\[= \cos{120{^\circ}} = \cos(180{^\circ} - 60{^\circ}) =\]

\[= - \cos{60{^\circ}} = - \frac{1}{2}\]

\[5)\sin\frac{47\pi}{6} = \sin{7\frac{5\pi}{6}} =\]

\[= \sin\left( 8\pi - \frac{\pi}{6} \right) = \sin\left( - \frac{\pi}{6} \right) =\]

\[= - \sin\frac{\pi}{6} = - \frac{1}{2}\]

\[6)\ tg\frac{25\pi}{4} = tg\ 6\frac{\pi}{4} =\]

\[= \text{tg}\left( 6\pi + \frac{\pi}{4} \right) = tg\frac{\pi}{4} = 1\]

\[7)\ ctg\frac{27\pi}{4} = ctg\ 6\frac{3\pi}{4} =\]

\[= \text{ctg}\left( 7\pi - \frac{\pi}{4} \right) = ctg\left( - \frac{\pi}{4} \right) =\]

\[= - ctg\frac{\pi}{4} = - 1\]

\[8)\cos\frac{21\pi}{4} = \cos{5\frac{\pi}{4}} =\]

\[= \cos\left( 4\pi + \pi + \frac{\pi}{4} \right) =\]

\[= \cos\left( \pi + \frac{\pi}{4} \right) = - \cos\frac{\pi}{4} = - \frac{\sqrt{2}}{2}\]