\[{\text{Докажите тождество:}}\] $%\frac{{{{\left( {\sin t + \cos t} \right)}^2} - 1}}{{{\mathop{\rm tg}\nolimits} t - \sin t\cos t}} = 2{{\mathop{\rm ctg}\nolimits} ^2}t$%

$$\eqalign{ {\text{Докажите}}{\text{, что}} \hfill \\ \frac{{\sin \left( {{{80}^ \circ } + \alpha } \right)}}{{4\sin \left( {{{20}^ \circ } + \frac{\alpha }{4}} \right) \cdot \sin \left( {{{70}^ \circ } - \frac{\alpha }{4}} \right)}} = \cos \left( {{{40}^ \circ } + \frac{\alpha }{2}} \right). \hfill \\ } $$

$$\eqalign{ {\text{Вычислите:}} \hfill \\ \prod\limits_{k = 1}^{2018} {\cos \frac{{\pi k}}{{2019}}} \hfill \\ } $$

$$\eqalign{ {\text{Вычислите }}\operatorname{tg} \frac{\pi }{{12}}. } $$

$$\eqalign{ {\text{Докажите}}{\text{, что}} \hfill \\ \cos \frac{\pi }{{{2^n}}} = \frac{1}{2}\sqrt {2 + \sqrt {2 + \sqrt {2 + ... + \sqrt 2 } } } \hfill \\ (n - 1{\text{ знак корня}}{\text{, }}n \in \mathbb{N}) \hfill \\ } $$

$$\eqalign{ {\text{Вычислите:}} \hfill \\ \cos \frac{{2\pi }}{7} + \cos \frac{{4\pi }}{7} + \cos \frac{{6\pi }}{7} \hfill \\ } $$

$$\eqalign{ {\text{Докажите тождество:}} \hfill \\ \frac{{1 + \cos 4\alpha + \sin 4\alpha }}{{1 - \cos 4\alpha + \sin 4\alpha }} = \operatorname{ctg} 2\alpha \hfill \\ } $$

$$\eqalign{ {\text{Докажите тождество:}} \hfill \\ \frac{{{{\sin }^2}2\alpha + 4{{\sin }^2}\alpha - 4}}{{1 - 8{{\sin }^2}\alpha - \cos 4\alpha }} = \frac{1}{2}{\operatorname{ctg} ^4}\alpha . \hfill \\ } $$

$$\eqalign{ {\text{Докажите тождество:}} \hfill \\ \frac{{\sin \alpha + \cos \alpha }}{{{{\cos }^3}\alpha }} = {\operatorname{tg} ^3}\alpha + {\operatorname{tg} ^2}\alpha + \operatorname{tg} \alpha + 1 \hfill \\ } $$

$$\eqalign{ {\text{Вычислите:}} \hfill \\ \cos \frac{\pi }{5} - \cos \frac{{2\pi }}{5}. \hfill \\ } $$

$$\eqalign{ \lg \operatorname{tg} {1^ \circ } + \lg \operatorname{tg} {2^ \circ } + \lg \operatorname{tg} {3^ \circ } + ... + \lg \operatorname{tg} {89^ \circ } } $$

$$\eqalign{ {\text{Вычислите:}} \hfill \\ \frac{{96 \cdot \sin {{80}^ \circ } \cdot \sin {{65}^ \circ } \cdot \sin {{35}^ \circ }}}{{\sin {{20}^ \circ } + \sin {{50}^ \circ } + \sin {{110}^ \circ }}} \hfill \\ } $$

$$\eqalign{ {\text{Докажите тождество:}} \hfill \\ \frac{{\sin \alpha + \sin 3\alpha + \sin 5\alpha }}{{\cos \alpha + \cos 3\alpha + \cos 5\alpha }} = \operatorname{tg} 3\alpha \hfill \\ } $$

$%\begin{array}{l} {\text{Пусть }}\alpha ,\beta ,\gamma {\text{ - углы треугольника}}{\text{. Докажите}}{\text{, что}} \hfill \\ \operatorname{tg} \alpha + \operatorname{tg} \beta + \operatorname{tg} \gamma = \operatorname{tg} \alpha \operatorname{tg} \beta \operatorname{tg} \gamma . \hfill \\ \end{array} $%

\[\begin{array}{l} {\text{Докажите тождество:}} \hfill \\ \cos \frac{\pi }{7} \cdot \cos \frac{{2\pi }}{7} \cdot \cos \frac{{4\pi }}{7} = - \frac{1}{8}. \hfill \\ \end{array} \]

\[{\text{Вычислить: а) }}\cos \frac{\pi }{5} \cdot \cos \frac{{2\pi }}{5};{\text{ б) }}\cos \frac{\pi }{5} - \cos \frac{{2\pi }}{5}.\]

\[\begin{array}{l} {\text{Известно}}{\text{, что }}\operatorname{ctg} x = 3.{\text{ Найдите значение выражения}} \hfill \\ {\sin ^2}\left( {{{30}^ \circ } + x} \right) - {\sin ^2}\left( {{{45}^ \circ } - x} \right) + \cos {75^ \circ }\sin \left( {{{75}^ \circ } + 2x} \right). \hfill \\ \end{array}\]

\[{\text{Покажите}}{\text{, что }}\cos {20^ \circ }\cos {40^ \circ }\cos {80^ \circ } = \frac{1}{8}.\]

\[\sin \frac{\pi }{{16}}{\cos ^3}\frac{\pi }{{16}} - {\sin ^3}\frac{\pi }{{16}}\cos \frac{\pi }{{16}}\]

\[{\text{Вычислите: }}\frac{1}{{\sin {{20}^ \circ }}} - \frac{1}{{\operatorname{tg} {{40}^ \circ }}}.\]

\[\begin{array}{l} \sqrt[3]{{\cos \frac{{2\pi }}{7}}} + \sqrt[3]{{\cos \frac{{4\pi }}{7}}} + \sqrt[3]{{\cos \frac{{8\pi }}{7}}} = \sqrt[3]{{\frac{{5 - 3\sqrt[3]{7}}}{2}}} \hfill \\ \sqrt[3]{{\cos \frac{{2\pi }}{9}}} + \sqrt[3]{{\cos \frac{{4\pi }}{9}}} + \sqrt[3]{{\cos \frac{{8\pi }}{9}}} = \sqrt[3]{{\frac{{3\sqrt[3]{9} - 6}}{2}}} \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите}}{\text{, что}} \hfill \\ {\sin ^4}\frac{\pi }{{16}} + {\sin ^4}\frac{{3\pi }}{{16}} + {\sin ^4}\frac{{5\pi }}{{16}} + {\sin ^4}\frac{{7\pi }}{{16}} = \frac{3}{2}. \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Найдите уравнение 4 - й степени с целыми коэффициентами}}{\text{, корни}} \hfill \\ {\text{которого }}{x_1} = 4{\sin ^2}\frac{\pi }{{16}},{\text{ }}{x_2} = 4{\sin ^2}\frac{{3\pi }}{{16}},{\text{ }}{x_3} = 4{\sin ^2}\frac{{5\pi }}{{16}},{\text{ }}{x_4} = 4{\sin ^2}\frac{{7\pi }}{{16}}. \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите}}{\text{, что }}\forall k \in \mathbb{N}{\text{ число}} \hfill \\ {\sin ^{2k}}\frac{\pi }{{16}} + {\sin ^{2k}}\frac{{3\pi }}{{16}} + {\sin ^{2k}}\frac{{5\pi }}{{16}} + {\sin ^{2k}}\frac{{7\pi }}{{16}} \hfill \\ {\text{рационально}}{\text{.}} \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите}}{\text{, что число }}\sum\limits_{k = 1}^n {{{\operatorname{tg} }^{2m}}\left( {\frac{{\pi k}}{{2n + 1}}} \right)} {\text{ целое}}{\text{.}} \hfill \\ m,n \in \mathbb{N} \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите тождество:}} \hfill \\ \sin \left( {\frac{1}{2}\arcsin \frac{1}{3}} \right) = \frac{{\sqrt 3 \cdot \left( {2 - \sqrt 2 } \right)}}{6}. \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите}}{\text{, что}} \hfill \\ \frac{\pi }{4} = 4\arctan \frac{1}{5} - \arctan \frac{1}{{239}}. \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите тождество:}} \hfill \\ \sqrt {\cos \frac{\pi }{5}} + \sqrt {\cos \frac{{2\pi }}{5}} = \frac{{\sqrt {4 + 2\sqrt 5 } }}{2}. \hfill \\ \end{array}\]

\[{\text{Вычислите: }}\sqrt[3]{{{{\operatorname{tg} }^{10}}\frac{\pi }{5}}} + \sqrt[3]{{{{\operatorname{tg} }^{10}}\frac{{2\pi }}{5}}}.\]

\[\begin{array}{l} {\text{Докажите тождество:}} \hfill \\ {\text{co}}{{\text{s}}^2}\frac{\pi }{7} + {\text{co}}{{\text{s}}^2}\frac{{2\pi }}{7} + {\text{co}}{{\text{s}}^2}\frac{{3\pi }}{7} = \frac{5}{4}. \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите тождество:}} \hfill \\ {\text{co}}{{\text{s}}^3}\frac{\pi }{7} - {\text{co}}{{\text{s}}^3}\frac{{2\pi }}{7} + {\text{co}}{{\text{s}}^3}\frac{{3\pi }}{7} = \frac{1}{2}. \hfill \\ \end{array}\]

\[\sqrt[3]{{{{\cos }^2}\frac{\pi }{7}}} + \sqrt[3]{{{{\cos }^2}\frac{{2\pi }}{7}}} + \sqrt[3]{{{{\cos }^2}\frac{{3\pi }}{7}}} = \sqrt[3]{{\frac{{6\sqrt[3]{7} + 3{{\sqrt[3]{7}}^2} + 11}}{4}}}.\]

\[\begin{array}{l} {\text{Докажите тождества:}} \hfill \\ {\text{а) }}\sum\limits_{k = 1}^3 {\sqrt[3]{{{{\left( {\frac{{\cos \frac{{\pi k}}{7}}}{{\cos \frac{{2\pi k}}{7}}}} \right)}^2}}}} = \sqrt[3]{{49}};{\text{ б) }}\sum\limits_{k = 1}^3 {\sqrt[3]{{{{\left( {\frac{{\cos \frac{{2\pi k}}{7}}}{{\cos \frac{{\pi k}}{7}}}} \right)}^2}}}} = 2\sqrt[3]{7}; \hfill \\ {\text{в) }}\sum\limits_{k = 1}^4 {\sqrt[3]{{{{\left( {\frac{{\cos \frac{{\pi k}}{9}}}{{\cos \frac{{2\pi k}}{9}}}} \right)}^2}}}} = 2\sqrt[3]{9} + 1;{\text{ г) }}\sum\limits_{k = 1}^4 {\sqrt[3]{{{{\left( {\frac{{\cos \frac{{2\pi k}}{9}}}{{\cos \frac{{\pi k}}{9}}}} \right)}^2}}}} = 3\sqrt[3]{3} + 1. \hfill \\ \end{array}\]

\[{\text{Докажите}}{\text{, что }}2\cos \frac{{2\pi }}{9}{\text{ является корнем уравнения }}{x^3} - 3x + 1 = 0.\]

\[\begin{array}{l} {\text{Докажите тождество:}} \hfill \\ \sqrt {\frac{{1 + \cos \frac{\pi }{8}}}{{1 - \cos \frac{\pi }{8}}}} + \sqrt {\frac{{1 - \cos \frac{\pi }{8}}}{{1 + \cos \frac{\pi }{8}}}} = 2\sqrt {4 + 2\sqrt 2 } . \hfill \\ \end{array}\]

$$\begin{array}{l} {\text{Пусть }}n \in \mathbb{N}.{\text{ Докажите формулу:}} \hfill \\ \operatorname{tg} \left( {n\operatorname{arctg} x} \right) = \frac{{\frac{1}{{2i}}\left( {{{\left( {1 + x \cdot i} \right)}^n} - {{\left( {1 - x \cdot i} \right)}^n}} \right)}}{{\frac{1}{2}\left( {{{\left( {1 + x \cdot i} \right)}^n} + {{\left( {1 - x \cdot i} \right)}^n}} \right)}}. \hfill \\ \end{array}$$

$$\sqrt[3]{{{{\left( {\frac{{\operatorname{tg} \frac{{2\pi }}{7}}}{{\operatorname{tg} \frac{\pi }{7}}}} \right)}^2}}} + \sqrt[3]{{{{\left( {\frac{{\operatorname{tg} \frac{{4\pi }}{7}}}{{\operatorname{tg} \frac{{2\pi }}{7}}}} \right)}^2}}} + \sqrt[3]{{{{\left( {\frac{{\operatorname{tg} \frac{{6\pi }}{7}}}{{\operatorname{tg} \frac{{3\pi }}{7}}}} \right)}^2}}} = \frac{4}{{\sqrt[3]{{3\sqrt[3]{7} - 5}}}}$$

\[\begin{array}{l} {\text{Докажите тождества:}} \hfill \\ {\text{а) }}\prod\limits_{k = 1}^n {\sin \frac{{\pi k}}{{2n + 1}}} = \frac{{\sqrt {2n + 1} }}{{{2^n}}}; \hfill \\ {\text{б) }}\prod\limits_{k = 1}^{n - 1} {\sin \frac{{\pi k}}{{2n}}} = \frac{{\sqrt n }}{{{2^{n - 1}}}}; \hfill \\ {\text{в) }}\prod\limits_{k = 1}^n {\cos \frac{{\pi k}}{{2n + 1}}} = \frac{1}{{{2^n}}}; \hfill \\ {\text{г) }}\prod\limits_{k = 1}^{n - 1} {\cos \frac{{\pi k}}{{2n}}} = \frac{{\sqrt n }}{{{2^{n - 1}}}}. \hfill \\ \end{array}\]

\[\begin{array}{l} {\text{Докажите}}{\text{, что}} \hfill \\ \prod\limits_{k = 1}^{n - 1} {\sin \frac{{\pi k}}{n}} = \frac{n}{{{2^{n - 1}}}}. \hfill \\ \end{array}\]

\[{\text{Докажите}}{\text{, что }}\frac{1}{{{{\operatorname{tg} }^4}\frac{\pi }{9}}} + \frac{1}{{{{\operatorname{tg} }^4}\frac{{2\pi }}{9}}} + \frac{1}{{{{\operatorname{tg} }^4}\frac{{4\pi }}{9}}} = 59.\]

\[{\text{Докажите}}{\text{, что }}\sin \left( {\frac{1}{4}\arcsin \frac{3}{4}} \right) = \frac{{\sqrt {6 - 2\sqrt 7 } }}{4}.\]

\[{\text{Find value of sin1}}{{\text{8}}^ \circ }.\]

\[\begin{array}{l} {\text{Prove the identity:}} \hfill \\ \sum\limits_{k = 1}^n {\left( {{{\sin }^{2m}}\left( {\frac{{\pi k}}{{2n}}} \right) + {{\cos }^{2m}}\left( {\frac{{\pi k}}{{2n}}} \right)} \right)} = 2n \cdot \frac{{\left( {2m - 1} \right)!!}}{{\left( {2m} \right)!!}}. \hfill \\ m,n \in \mathbb{N} \hfill \\ \end{array} \]

$$\begin{array}{l}\cos \frac{\pi }{{17}} \cdot \cos \frac{{4\pi }}{{17}} = \frac{{1 + \sqrt {17} + \sqrt {34 + 2\sqrt {17} } }}{{16}} \hfill \\ \frac{1}{{\cos \frac{\pi }{{17}}}} + \frac{1}{{\cos \frac{{2\pi }}{{17}}}} - \frac{1}{{\cos \frac{{4\pi }}{{17}}}} + \frac{1}{{\cos \frac{{8\pi }}{{17}}}} = 2\sqrt {17 + 4\sqrt {17} }\end{array} $$

$$\operatorname{tg} \left( {\frac{1}{4}\operatorname{arctg} \frac{4}{3}} \right) = \sqrt 5 - 2.$$ $$\operatorname{tg} \left( {\frac{1}{3}\operatorname{arctg} \frac{{\sqrt 5 }}{7}} \right) = \frac{{3\sqrt 3 - 2\sqrt 5 }}{7}$$ $$\operatorname{tg} \left( {\frac{1}{4}\operatorname{arctg} \frac{{\sqrt 5 }}{2}} \right) = \sqrt 6 - \sqrt 5 $$ $$\operatorname{tg} \left( {\frac{1}{4}\operatorname{arctg} \frac{1}{8}} \right) = \frac{{\sqrt 7 }}{7}$$