Question

помогите пожалуйста!!
Найти sin a и tg a , если cos a = - 12/13 (дробь)​

Answer 1

$\it \large \displaystyle 1)\sin^2\alpha +\cos^2\alpha =1\Longrightarrow \sin\alpha =\pm \sqrt{1-\cos^2\alpha } \Rightarrow \sin\alpha =\pm \sqrt{1-\bigg(-\frac{12}{13}\bigg)^2 }=\pm \sqrt{1-\frac{144}{169} }=\pm\sqrt{\frac{25}{169} } =\boxed{\Huge \bf \pm \frac{5}{13} } ;\\\tt \large \displaystyle 2)\sin^2\alpha +cos^2\alpha =1\hookrightarrow tg\alpha=\frac{\sin\alpha }{\cos\alpha } =\pm \dfrac{\dfrac{5}{13} }{-\dfrac{12}{13} } =\pm \frac{5}{-12} =\boxed{\bf \Huge \pm -\frac{5}{12} } .$

$\Huge \bf \cdot \cdot\\\Huge \smile$