Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 502

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\[1)\ 2\sin{75{^\circ}} \bullet \cos{75{^\circ}} =\]

\[= \sin(2 \bullet 75{^\circ}) = \sin{150{^\circ}} =\]

\[= \sin{30{^\circ}} = \frac{1}{2}\]

\[2)\cos^{2}{75{^\circ}} - \sin^{2}{75{^\circ}} =\]

\[= \cos{(2 \bullet 75{^\circ}}) = \cos{150{^\circ}} =\]

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

\[3)\ \frac{6\ tg\ 75{^\circ}}{1 - tg^{2}\ 75{^\circ}\ } =\]

\[= 3 \bullet \frac{2\ tg\ 75{^\circ}}{1 - tg^{2}\ 75{^\circ}} =\]

\[= 3 \bullet tg\ (2 \bullet 75{^\circ}) = 3 \bullet tg\ 150{^\circ} =\]

\[= - 3\ tg\ 30{^\circ} = - 3 \bullet \frac{\sqrt{3}}{3} = - \sqrt{3}\]

\[4)\ \frac{tg^{2}\ \left( 22{^\circ}{\ 30}^{'} \right) - 1}{\text{tg\ }\left( 22{^\circ}{\ 30}^{'} \right)} =\]

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

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

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

\[= - 2 \bullet (tg\ 45{^\circ})^{- 1} = - 2 \bullet 1^{- 1} =\]

\[= - 2\]