Question

if cos x= (-5/13) and tan x <0, then find sin 2x

Answer 1

Conditions

if $\cos x = (-5/13)$ and $\tan x < 0$, then find $\sin 2x$

Solution

$$\cos x = - \frac {5}{13}$$ $$x = \pm \arccos \frac {5}{13} + 2\pi n$$ $$\tan x < 0$$ $$\tan x = \frac {\sin x}{\cos x} = - \frac {13 \sin x}{5}, \text{ then } \sin x > 0$$ $$\sin^2 x + \cos^2 x = 1$$ $$\sin^2 x = 1 - \frac {25}{169} = \frac {144}{169}$$ $$\sin x = + \frac {12}{13}$$ $$\sin 2x = 2 \sin x \cos x = - \frac {10}{13} \sin x = - \frac {120}{169}$$