ГДЗ по алгебре 11 класс Никольский Параграф 3. Обратные функции Задание 19

\[\boxed{\mathbf{19}\mathbf{.}}\]

\[\textbf{а)}\cos\left( \arccos x \right) = x;\]

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

\[x \in \lbrack - 1;1\rbrack.\]

\[\cos a = a;\ \ |a| \leq 1:\]

\[\cos\left( \arccos a \right) = a;\]

\[\cos\left( \arccos x \right) = x.\]

\[Что\ и\ требовалось\ доказать.\]

\[\sin{(a)} = \sqrt{1 - \cos^{2}a}\]

\[\sin\left( \arccos x \right) =\]

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

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

\[Что\ и\ требовалось\ доказать.\]

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

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

\[\cos a = a;\ \ |a| \leq 1:\]

\[\cos\left( \arccos a \right) = a;\]

\[\cos\left( \arccos x \right) = x.\]

\[\text{ctg}(a) = \frac{\cos a}{\sin a};\ \ a = \arccos x;\ \]

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

\[\text{ctg}\left( \arccos x \right) = \frac{\cos\left( \arccos x \right)}{\sin\left( \arccos x \right)} =\]

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

\[Что\ и\ требовалось\ доказать.\]

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

\[x \in \lbrack - 1;0) \cup (0;1\rbrack.\]

\[tga = \frac{\sin a}{\cos a};\ \ a = \arccos x;\ \ \]

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

\[\cos\left( \arccos a \right) = a:\]

\[\text{tg}\left( \arccos x \right) = \frac{\sin\left( \arccos x \right)}{\cos\left( \arccos x \right)} =\]

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

\[Что\ и\ требовалось\ доказать.\]

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

\[\text{tg}(a) = a\ (из\ определения);\]

\[\text{tg}\left( \text{arctg}(a) \right) = a;\]

\[\text{tg}\left( \text{arctg}(x) \right) = x.\]

\[Что\ и\ требовалось\ доказать.\]

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

\[\ x \in R\]

\[1 + tg^{2}(a) = \frac{1}{\cos^{2}a}\]

\[\cos^{2}a = \frac{1}{1 + tg^{2}a};\]

\[a = arctg(x);\]

\[\text{tg}(a) = x.\]

\[\cos^{2}a = \frac{1}{1 + x^{2}}\]

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

\[Что\ и\ требовалось\ доказать.\]

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

\[x \in R\]

\[tga = \frac{\sin a}{\cos a}\]

\[\sin a = tga \cdot \cos a\]

\[\sin\left( \text{arctg}(x) \right) =\]

\[= tg\left( \text{arctg\ x} \right) \cdot \cos\left( \text{arctgx} \right) =\]

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

\[Что\ и\ требовалось\ доказать.\]

\[\textbf{д)}\ ctg\left( \text{arctg}(x) \right) = \frac{1}{x};\ \ x \neq 0\]

\[ctg\ a = \frac{1}{\text{tga}};\ \ a = arctgx;\]

\[\text{tg}(a) = x:\]

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

\[Что\ и\ требовалось\ доказать.\]

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

\[\text{ctg}\left( \text{arccot}a \right) = a - из\]

\[\ определения:\]

\[\text{ctg}\left( \text{arccot}x \right) = x.\]

\[Что\ и\ требовалось\ доказать.\]

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

\[1 + ctg^{2}a = \frac{1}{\sin^{2}a}\]

\[\sin^{2}a = \frac{1}{1 + ctg^{2}a};\ \ \]

\[a = \text{arccot}x;\ \ ctg\ a = x:\]

\[\sin^{2}a = \frac{1}{1 + x^{2}}\]

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

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

\[Что\ и\ требовалось\ доказать.\]

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

\[\cos a = ctg\ a \cdot \sin a\]

\[a = \text{arccot}x:\]

\[\cos\left( \text{arccot}x \right) =\]

\[= ctg\left( \text{arccot}x \right) \cdot \sin\left( \text{arccot}x \right) =\]

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

\[Что\ и\ требовалось\ доказать.\]

\[\textbf{ж)}\ tg\left( \text{arccot}x \right) = \frac{1}{x}\]

\[tga = \frac{1}{\text{ctga}};\ \ a = \text{arccot}x;\ \ \]

\[\text{ctg}\left( \text{arccot}x \right) = x:\]

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

\[Что\ и\ требовалось\ доказать.\]