Задание 12.8

Question

$\DeclareMathOperator{	g}{tg}$
Найдите значения тригонометрических функций аргумента $\alpha$, если:
ewline
$$1)\quad \cos \alpha = \frac{12}{13} 	ext{ и } 0 < \alpha < \frac{\pi}{2}; \qquad 3) \quad 	g \alpha = -\frac{1}{3} 	ext{ и }\frac{3\pi}{2} < \alpha < 2\pi.$$
$$2) \sin \alpha = -\frac{\sqrt{3}}{4} 	ext{ и } \pi < \alpha < \frac{3\pi}{2};$$

Gimli · Accepted Answer

$\DeclareMathOperator{	g}{tg}$
$\DeclareMathOperator{\ctg}{ctg}$
\begin{enumerate}
\item $$\bbox[13px,border:3px solid blue]{\sin^2\alpha + \cos^2\alpha = 1}.$$
$$\cos \alpha = \frac{12}{13}\Longrightarrow \sin \alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}}= \frac{5}{13} \Longrightarrow 	g \alpha= \frac{\sin\alpha}{\cos\alpha} = \frac{5\cdot 13}{13\cdot 12} = \frac{5}{12}\Longrightarrow $$ $$\ctg \alpha = \frac{12}{5} = 2\frac{2}{5}.$$
\item $$\sin \alpha = -\frac{\sqrt{3}}{4}\Longrightarrow \cos \alpha = -\sqrt{1 - \sin^2\alpha} = -\sqrt{1 - \frac{3}{16}} = -\sqrt{\frac{13}{16}}\Longrightarrow 	g \alpha = \frac{\sin \alpha}{\cos\alpha} = \sqrt{\frac{3}{13}}\Longrightarrow \ctg \alpha = \sqrt{\frac{13}{3}}$$
\item $$\bbox[8px,border:3px solid blue]{1 + 	g^2\alpha= \frac{1}{\cos^2\alpha}}$$
$$	g \alpha = -\frac{1}{3} \Longrightarrow \bbox[8px,border:3px solid red]{\ctg \alpha = -3}\Longrightarrow 1 + \frac{1}{9} = \frac{1}{\cos^2\alpha}\Longrightarrow \frac{10}{9} = \frac{1}{\cos^2\alpha}\Longrightarrow 10\cos^2\alpha = 9\Longrightarrow \cos^2\alpha = \frac{9}{10}\Longrightarrow $$ $$\bbox[8px,border:3px solid red]{\cos\alpha = \frac{3}{\sqrt{10}}}	ext{(значение синуса могло быть и отрицательным, но из за значения альфа он положителен)}\Rightarrow$$ $$\Longrightarrow \bbox[8px,border:3px solid red]{\sin \alpha = -\frac{1}{\sqrt{10}}.}$$
\end{enumerate}

Задание 12.8

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Задание 12.8

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