Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 483

\[\boxed{\mathbf{483.}}\]

\[1)\cos\left( \frac{\pi}{3} + a \right)\]

\[\sin a = \frac{1}{\sqrt{3}}\text{\ \ }и\ \ 0 < a < \frac{\pi}{2}\]

\[\ в\ I\ четверти:\]

\[\cos a = \sqrt{1 - \sin^{2}a} =\]

\[= \sqrt{1 - \left( \frac{1}{\sqrt{3}} \right)^{2}} = \sqrt{\frac{3}{3} - \frac{1}{3}} = \sqrt{\frac{2}{3}}\]

\[Получаем:\]

\[\cos\left( \frac{\pi}{3} + a \right) =\]

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

\[\cos\left( \frac{\pi}{3} + a \right) = \frac{1}{2} \bullet \sqrt{\frac{2}{3}} - \frac{\sqrt{3}}{2} \bullet \frac{1}{\sqrt{3}} =\]

\[= \frac{1}{\sqrt{2} \bullet \sqrt{3}} - \frac{1}{2} = \frac{1}{\sqrt{6}} - \frac{1}{2}\]

\[Ответ:\ \ \frac{1}{\sqrt{6}} - \frac{1}{2}.\]

\[2)\cos\left( a - \frac{\pi}{4} \right)\]

\[\cos a = - \frac{1}{3}\text{\ \ }и\ \ \frac{\pi}{2} < a < \pi\]

\[во\text{\ II\ }четверти:\]

\[\sin a = \sqrt{1 - \cos^{2}a} =\]

\[= \sqrt{1 - \left( - \frac{1}{3} \right)^{2}} = \sqrt{\frac{9}{9} - \frac{1}{9}} =\]

\[= \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\]

\[Получаем:\]

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

\[= \cos a \bullet \cos\frac{\pi}{4} + \sin a \bullet \sin\frac{\pi}{4}\]

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

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

\[= - \frac{\sqrt{2}}{6} + \frac{2 \bullet 2}{6} = \frac{4 - \sqrt{2}}{6}\]

\[Ответ:\ \ \frac{4 - \sqrt{2}}{6}.\]