Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1276

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\[\alpha + \beta = \varphi\]

\[\frac{a}{b} = k;\ \ \ \ \ \frac{\sin\alpha}{a} = \frac{\sin\beta}{b}\]

\[\frac{\sin\alpha}{\sin\beta} = k\]

\[\frac{a - b}{a + b} = \frac{\text{tg}\frac{\alpha - \beta}{2}}{\text{tg}\frac{\alpha + \beta}{2}} = \frac{k - 1}{k + 1} =\]

\[= \frac{\text{tg}\frac{\alpha - \beta}{2}}{\text{tg}\frac{\varphi}{2}}\]

\[\text{tg}\frac{\alpha - \beta}{2} = \frac{(k - 1)tg\frac{\varphi}{2}}{k + 1}\]

\[\frac{\alpha - \beta}{2} = arctg\ \frac{(k - 1)tg\frac{\varphi}{2}}{k + 1}\]

\[\left\{ \begin{matrix} \alpha - \beta = 2arctg\ \frac{(k - 1)\text{tg}\frac{\varphi}{2}}{k + 1} \\ \alpha + \beta = \varphi\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 2\alpha = \varphi + 2arctg\ \frac{(k - 1)\text{tg}\frac{\varphi}{2}}{k + 1} \\ 2\beta = \varphi - 2arctg\ \frac{(k - 1)\text{tg}\frac{\varphi}{2}}{k + 1} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \alpha = \frac{\varphi}{2} + arctg\ \frac{(k - 1)\text{tg}\frac{\varphi}{2}}{k + 1} \\ \beta = \frac{\varphi}{2} - arctg\ \frac{(k - 1)\text{tg}\frac{\varphi}{2}}{k + 1} \\ \end{matrix} \right.\ \]