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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 1067
Задание 1067
\[\boxed{\mathbf{1067}\mathbf{.}}\]
\[1)\sin{15{^\circ}}:\]
\[\sin^{2}{15{^\circ}} = \frac{1 - \cos(2 \bullet 15{^\circ})}{2} =\]
\[= \frac{1 - \cos{30{^\circ}}}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} =\]
\[= \frac{1}{2}\left( \frac{2}{2} - \frac{\sqrt{3}}{2} \right) = \frac{2 - \sqrt{3}}{4}\]
\[\sin{15{^\circ}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}\]
\[2)\cos{15{^\circ}}:\]
\[\cos^{2}{15{^\circ}} = \frac{1 + \cos(2 \bullet 15{^\circ})}{2} =\]
\[= \frac{1 + \cos{30{^\circ}}}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} =\]
\[= \frac{1}{2}\left( \frac{2}{2} + \frac{\sqrt{3}}{2} \right) = \frac{2 + \sqrt{3}}{4}\]
\[\cos{15{^\circ}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}\]
\[3)\ tg\ 22{^\circ}\ 30^{'}:\]
\[tg^{2}\ 22{^\circ}\ 30^{'} =\]
\[= \frac{1 - \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)}{1 + \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)} =\]
\[= \frac{1 - \cos{45{^\circ}}}{1 + \cos{45{^\circ}}} =\]
\[= \frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} = \frac{\frac{1}{\sqrt{2}}\left( \sqrt{2} - 1 \right)}{\frac{1}{\sqrt{2}}\left( \sqrt{2} + 1 \right)}\]
\[tg^{2}\ 22{^\circ}\ 30^{'} =\]
\[= \frac{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} - 1 \right)}{\left( \sqrt{2} + 1 \right)\left( \sqrt{2} - 1 \right)} =\]
\[= \frac{\left( \sqrt{2} - 1 \right)^{2}}{2 - 1} = \left( \sqrt{2} - 1 \right)^{2}\]
\[tg\ 22{^\circ}\ 30^{'} = \sqrt{\left( \sqrt{2} - 1 \right)^{2}} =\]
\[= \sqrt{2} - 1\]
\[4)\ ctg\ 22{^\circ}\ 30^{'}:\]
\[\text{ct}g^{2}\ 22{^\circ}\ 30^{'} =\]
\[= \frac{1 + \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)}{1 - \cos\left( 2 \bullet 22{^\circ}\ 30^{'} \right)} =\]
\[= \frac{1 + \cos{45{^\circ}}}{1 - \cos{45{^\circ}}} =\]
\[= \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} = \frac{\frac{1}{\sqrt{2}}\left( \sqrt{2} + 1 \right)}{\frac{1}{\sqrt{2}}\left( \sqrt{2} - 1 \right)}\]
\[\text{ct}g^{2}\ 22{^\circ}\ 30^{'} =\]
\[= \frac{\left( \sqrt{2} + 1 \right)\left( \sqrt{2} + 1 \right)}{\left( \sqrt{2} - 1 \right)\left( \sqrt{2} + 1 \right)} =\]
\[= \frac{\left( \sqrt{2} + 1 \right)^{2}}{2 - 1} = \left( \sqrt{2} + 1 \right)^{2}\]
\[ctg\ 22{^\circ}\ 30^{'} = \sqrt{\left( \sqrt{2} + 1 \right)^{2}} =\]
\[= \sqrt{2} + 1\]
