ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 517

\[\boxed{\mathbf{517}\mathbf{.}}\]

\[1)\sin{15{^\circ}}:\]

\[\sin^{2}{15{^\circ}} = \frac{1 - \cos(2 \bullet 15{^\circ})}{2} =\]

\[= \frac{1 - \cos{30{^\circ}}}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} =\]

\[= \frac{1}{2}\left( \frac{2}{2} - \frac{\sqrt{3}}{2} \right) = \frac{2 - \sqrt{3}}{4}\]

\[\sin{15{^\circ}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}\]

\[2)\cos{15{^\circ}}:\]

\[\cos^{2}{15{^\circ}} = \frac{1 + \cos(2 \bullet 15{^\circ})}{2} =\]

\[= \frac{1 + \cos{30{^\circ}}}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} =\]

\[= \frac{1}{2}\left( \frac{2}{2} + \frac{\sqrt{3}}{2} \right) = \frac{2 + \sqrt{3}}{4}\]

\[\cos{15{^\circ}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}\]

\[3)\ tg\ 22{^\circ}\ 30^{'}:\]

\[tg^{2}\ 22{^\circ}\ 30^{'} =\]

\[= \frac{1 - \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)}{1 + \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)} =\]

\[= \frac{1 - \cos{45{^\circ}}}{1 + \cos{45{^\circ}}} = \frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} =\]

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

\[tg^{2}\ 22{^\circ}\ 30^{'} =\]

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

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

\[tg\ 22{^\circ}\ 30^{'} = \sqrt{\left( \sqrt{2} - 1 \right)^{2}} =\]

\[= \sqrt{2} - 1\]

\[4)\ ctg\ 22{^\circ}\ 30^{'}:\]

\[\text{ct}g^{2}\ 22{^\circ}\ 30^{'} =\]

\[= \frac{1 + \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)}{1 - \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)} =\]

\[= \frac{1 + \cos{45{^\circ}}}{1 - \cos{45{^\circ}}} = \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} =\]

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

\[\text{ct}g^{2}\ 22{^\circ}\ 30^{'} =\]

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

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

\[ctg\ 22{^\circ}\ 30^{'} = \sqrt{\left( \sqrt{2} + 1 \right)^{2}} =\]

\[= \sqrt{2} + 1\]