Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 498

\[\boxed{\mathbf{498.}}\]

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

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

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

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

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

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

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

\[4)\sin\frac{4\pi}{3} =\]

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

\[= 2\sin\frac{2\pi}{3} \bullet \cos\frac{2\pi}{3}\]

\[5)\cos\frac{5\pi}{3} =\]

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

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