Решебник по алгебре 11 класс Никольский Параграф 3. Обратные функции Задание 20

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\[\textbf{а)}\ y = \cos\left( \arccos x \right);\ \ \lbrack - 1;1\rbrack\]

\[y = x;\ \ \lbrack - 1;1\rbrack\]

\[\textbf{б)}\ y = \sin\left( \arccos x \right);\lbrack - 1;1\rbrack\]

\[\sin\left( \arccos x \right) = \sqrt{1 - x^{2}}\]

\[y = \sqrt{1 - x^{2}}\]

\[y^{2} = 1 - x^{2}\]

\[y^{2} + x^{2} = 1;\ \ y \in \lbrack 0;1\rbrack\]

\[\textbf{в)}\ y = tg\ \left( \arccos x \right);\ \ \]

\[D(f) = \lbrack - 1;0) \cup (0;1\rbrack.\]

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

\[y = \frac{\sqrt{1 - x^{2}}}{x} = \sqrt{\frac{1}{x^{2}} - 1}\]

\[\textbf{г)}\ y = ctg\ \left( \arccos x \right)\]

\[\text{ctg}\left( \arccos x \right) = \frac{x}{\sqrt{1 - x^{2}}};\ \ \ \]

\[x \in ( - 1;1)\]

\[y = \frac{x}{\sqrt{1 - x^{2}}}\]

\[\textbf{д)}\ y = tg\ (\arctan x)\]

\[\text{tg}\left( \text{arctg}(x) \right) = x;\ \ x \in R\]

\[y = x\]

\[\textbf{е)}\ y = ctg\ \left( \arctan x \right);\ \ x \neq 0\]

\[\text{ctg\ }\left( \arctan x \right) = \frac{1}{x}\]

\[y = \frac{1}{x}\]

\[\textbf{ж)}\ y = \sin{(\arctan x)}\]

\[\sin\left( \text{arctg}(x) \right) = \frac{x}{\sqrt{1 + x^{2}}};\ \ x \in R\]

\[y = \frac{x}{\sqrt{1 + x^{2}}}\]

\[\textbf{з)}\ y = \cos{(\arctan x)}\]

\[\cos\left( \text{arctg}(x) \right) = \frac{1}{\sqrt{1 + x^{2}}};\ \ x \in R\]

\[y = \frac{1}{\sqrt{1 + x^{2}}}\]

\[\textbf{и)}\ y = ctg\ (\text{arccot}x)\]

\[\text{ctg}\left( \text{arccot}x \right) = x;\ x \in R\]

\[y = x\]

\[к)\ y = tg\ (\text{arccot}x)\]

\[\text{tg\ }\left( \text{arccot}x \right) = \frac{1}{x};\ \ \ x \neq 0\]

\[y = \frac{1}{x}\]

\[л)\ y = \sin{(\text{arccot}x)}\]

\[\sin\left( \text{arccot}x \right) = \frac{1}{\sqrt{1 + x^{2}}};\ \ x \in R\]

\[y = \frac{1}{\sqrt{1 + x^{2}}}\]

\[м)\ y = \cos{(\text{arccot}x)}\]

\[\cos{(\text{arccot}x)} = \frac{x}{\sqrt{1 + x^{2}}};\ \ x \in R\]

\[y = \frac{x}{\sqrt{1 + x^{2}}}\]