Question

відомо, що
$\cot( \alpha ) = - 3$
і a - кут 4 чверті. Знайдіть
$\sin( \alpha )$
$\cos( \alpha )$
$\tan( \alpha )$
​

Answer 1

Ответ:

$\displaystyle sin\alpha=-\frac{\sqrt{10}}{10}\\\\cos\alpha=\frac{3\sqrt{10}}{10}\\\\tan\alpha=- \frac{ 1 }{ 3 }$

Пошаговое объяснение:

$\displaystyle cot\alpha=\frac{cos\alpha}{sin\alpha}\\\\ \frac{cos\alpha}{sin\alpha}=-3\\\\ cos\alpha=-3sin\alpha$

$\displaystyle sin^2\alpha+cos^2\alpha=1\\\\ sin^2\alpha+\left( -3sin\alpha\right)^2 =1\\\\ sin^2\alpha+9sin^2\alpha=1\\\\10sin^2\alpha=1\ \ \ \ \ \ |:10\\\\ sin^2\alpha= \frac{ 1 }{ 10 } \\\\ sin\alpha=-\sqrt{ \frac{ 1 }{ 10 } }\\\\ sin\alpha=-\frac{\sqrt{10}}{10}$

$\displaystyle cos\alpha=-3sin\alpha\\\\ cos\alpha=-3\cdot\left( -\frac{\sqrt{10}}{10} \right) \\\\ cos\alpha=\frac{3\sqrt{10}}{10}$

$\displaystyle tan\alpha=\frac{1}{cot\alpha}\\\\ tan\alpha=\frac{1}{-3}\\\\ tan\alpha=- \frac{ 1 }{ 3 }$