ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 1045

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\[1)\sin{48{^\circ}} = 2\sin\frac{48{^\circ}}{2} \bullet \cos\frac{48{^\circ}}{2} =\]

\[= 2\sin{24{^\circ}} \bullet \cos{24{^\circ}}\]

\[2)\cos{164{^\circ}} = \cos^{2}\frac{164{^\circ}}{2} -\]

\[- \sin^{2}\frac{164{^\circ}}{2} = \cos^{2}{82{^\circ}} -\]

\[- \sin^{2}{82{^\circ}}\]

\[3)\ tg\ 92{^\circ} = \frac{2\ tg\frac{92{^\circ}}{2}}{1 - tg^{2}\frac{92{^\circ}}{2}} =\]

\[= \frac{2\ tg\ 46{^\circ}}{1 - tg^{2}\ 46{^\circ}}\]

\[4)\sin\frac{4\pi}{3} = 2\sin\left( \frac{1}{2} \bullet \frac{4\pi}{3} \right) \bullet\]

\[{\bullet \cos}\left( \frac{1}{2} \bullet \frac{4\pi}{3} \right) = 2\sin\frac{2\pi}{3} \bullet \cos\frac{2\pi}{3}\]

\[5)\cos\frac{5\pi}{3} = \cos^{2}\left( \frac{1}{2} \bullet \frac{5\pi}{3} \right) -\]

\[- \sin^{2}\left( \frac{1}{2} \bullet \frac{5\pi}{3} \right) = \cos^{2}\frac{5\pi}{6} -\]

\[- \sin^{2}\frac{5\pi}{6}\]