Решебник по алгебре 11 класс Никольский Параграф 3. Обратные функции Задание 22

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Год:2020-2021-2022
Тип:учебник

Задание 22

\[\boxed{\mathbf{22}\mathbf{.}}\]

\[\cos(a + \beta + \gamma) =\]

\[= \cos{(a +}\beta) \cdot \cos\gamma -\]

\[- \sin(a + \beta) \cdot \sin\gamma =\]

\[= \left( \cos a \cdot \cos\beta - \sin a \cdot \sin\beta \right) \cdot\]

\[\cdot \cos\gamma -\]

\[- \left( \sin a \cdot \cos\beta + \cos a \cdot \sin\beta \right) \cdot\]

\[\cdot \sin\gamma;\]

\[\cos a = \cos\left( \arccos\frac{4}{5} \right) = \frac{4}{5};\ \]

\[\cos\beta = \cos\left( \arccos\frac{12}{13} \right) = \frac{12}{13};\]

\[\cos\gamma = \cos\left( \arccos\frac{3}{5} \right) = \frac{3}{5};\]

\[\sin a = \sqrt{1 - \left( \frac{4}{5} \right)^{2}} = \sqrt{1 - \frac{16}{25}} =\]

\[= \sqrt{\frac{9}{25}} = \frac{3}{5};\]

\[\sin\beta = \sqrt{1 - \left( \frac{12}{13} \right)^{2}} = \sqrt{1 - \frac{144}{169}} =\]

\[= \sqrt{\frac{25}{169}} = \frac{5}{13};\]

\[\sin\gamma = \sqrt{1 - \left( \frac{3}{5} \right)^{2}} = \sqrt{1 - \frac{9}{25}} =\]

\[= \sqrt{\frac{16}{25}} = \frac{4}{5}.\]

\[\left( \cos a \cdot \cos\beta - \sin a \cdot \sin\beta \right) \cdot\]

\[\cdot \cos\gamma -\]

\[- \left( \sin a \cdot \cos\beta + \cos a \cdot \sin\beta \right) \cdot\]

\[\cdot \sin\gamma =\]

\[= \left( \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} \right) \cdot \frac{3}{5} -\]

\[- \left( \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13} \right) \cdot \frac{4}{5} =\]

\[= \left( \frac{48}{65} - \frac{15}{65} \right) \cdot \frac{3}{5} - \left( \frac{36}{65} + \frac{20}{65} \right) \cdot\]

\[\cdot \frac{4}{5} = \frac{33}{65} \cdot \frac{3}{5} - \frac{56}{65} \cdot \frac{4}{5} =\]

\[= \frac{99}{325} - \frac{224}{325} = - \frac{125}{325} = - \frac{5}{13}.\]

\[Ответ:\ - \frac{5}{13}.\]

\[\textbf{б)}\sin\left( \arcsin\frac{3}{5} + \arcsin\frac{5}{13} + \arcsin\frac{4}{5} \right) =\]

\[= \sin(a + \beta + \gamma)\]

\[\sin(a + \beta + \gamma) = \sin(a + \beta) \cdot\]

\[\cdot \cos\gamma + \cos(a + \beta) \cdot \sin\gamma =\]

\[= \left( \sin a \cdot \cos\beta + \cos a \cdot \sin\beta \right) \cdot\]

\[\cdot \cos\gamma +\]

\[+ \left( \cos a \cdot \cos\beta - \sin a \cdot \sin\beta \right) \cdot\]

\[\cdot \sin\gamma;\]

\[\sin a = \sin\left( \arcsin\frac{3}{5} \right) = \frac{3}{5};\]

\[\sin\beta = \sin\left( \arcsin\frac{5}{13} \right) = \frac{5}{13};\]

\[\sin\gamma = \sin\left( \arcsin\frac{4}{5} \right) = \frac{4}{5};\]

\[\cos a = \sqrt{1 - \left( \frac{3}{5} \right)^{2}} = \sqrt{1 - \frac{9}{25}} =\]

\[= \sqrt{\frac{16}{25}} = \frac{4}{5};\]

\[\cos\beta = \sqrt{1 - \left( \frac{5}{13} \right)^{2}} = \sqrt{1 - \frac{25}{169}} =\]

\[= \sqrt{\frac{144}{169}} = \frac{12}{13};\]

\[\cos\gamma = \sqrt{1 - \left( \frac{4}{5} \right)^{2}} = \sqrt{1 - \frac{16}{25}} =\]

\[= \sqrt{\frac{9}{25}} = \frac{3}{5}.\]

\[\left( \sin a \cdot \cos\beta + \cos a \cdot \sin\beta \right) \cdot\]

\[\cdot \cos\gamma +\]

\[+ \left( \cos a \cdot \cos\beta - \sin a \cdot \sin\beta \right) \cdot \sin\gamma =\]

\[= \left( \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13} \right) \cdot \frac{3}{5} +\]

\[+ \left( \frac{4}{5} \cdot \frac{12}{13} - \frac{3}{5} \cdot \frac{5}{13} \right) \cdot \frac{4}{5} =\]

\[= \left( \frac{36}{65} + \frac{20}{65} \right) \cdot \frac{3}{5} + \left( \frac{48}{65} - \frac{15}{65} \right) \cdot\]

\[\cdot \frac{4}{5} = \frac{56}{65} \cdot \frac{3}{5} + \frac{33}{65} \cdot \frac{4}{5} = \frac{168}{325} +\]

\[+ \frac{132}{325} = \frac{300}{325} = \frac{12}{13}.\]

\[Ответ:\frac{12}{13}.\]

\[\textbf{в)}\ ctg\left( \text{arctg}\frac{1}{3} + arctg\frac{1}{4} + arctg\frac{2}{9} \right) =\]

\[= ctg(a + \beta + \gamma)\]

\[\text{\ ctg}(a + \beta + \gamma) =\]

\[= \frac{- 1 + ctg(a + \beta) \cdot ctg\ \gamma}{\text{ctg}(a + \beta) + ctg\gamma} =\]

\[= \frac{- 1 + \frac{- 1 + ctga \cdot ctg\beta}{ctga + ctg\beta} \cdot ctg\gamma}{\frac{- 1 + ctga \cdot ctg\beta}{ctga + ctg\beta} + ctg\gamma};\]

\[ctg\ a = 1\ :\frac{1}{3} = 3;\]

\[ctg\ \beta = 1\ :\frac{1}{4} = 4;\]

\[ctg\ \gamma = 1\ :\frac{2}{9} = \frac{9}{2};\]

\[\frac{- 1 + \frac{- 1 + ctga \cdot ctg\beta}{ctga + ctg\beta} \cdot ctg\gamma}{\frac{- 1 + ctga \cdot ctg\beta}{ctga + ctg\beta} + ctg\gamma} =\]

\[= \frac{- 1 + \frac{- 1 + 3 \cdot 4}{3 + 4} \cdot \frac{9}{2}}{\frac{- 1 + 3 \cdot 4}{3 + 4} + \frac{9}{2}} =\]

\[= \frac{- 1 + \frac{11}{7} \cdot \frac{9}{2}}{\frac{11^{\backslash 2}}{7} + \frac{9^{\backslash 7}}{2}} = \frac{- 1^{\backslash 14} + \frac{99}{14}}{\frac{22 + 63}{14}} =\]

\[= \frac{85}{14}\ :\frac{85}{14} = 1.\]

\[Ответ:1.\]

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