# If cos A = -4/5 how do you find tan 2A?

##### 2 Answers
May 21, 2018

­±40/7 * 3/5 = ± 24/7

#### Explanation:

$\cos x = - \frac{4}{5}$

$\tan 2 x = \frac{\sin 2 x}{\cos 2 x} = \frac{2 \sin x \cos x}{{\cos}^{2} x - {\underbrace{{\sin}^{2} x}}_{1 - {\cos}^{2} x}} = \frac{- 2 \cdot \sin x \cdot \frac{4}{5}}{2 \cdot \frac{16}{25} - 1}$

± sin x = sqrt {1 - 16/25} = sqrt{{25-16}/25} = 3/5

$\tan 2 x = - \frac{8}{5} \cdot \sin x \cdot \frac{25}{32 - 25} = - \frac{40}{7} \sin x$

May 22, 2018

24/7

#### Explanation:

$\cos A = - \frac{4}{5}$.
A could be in Quadrant 2 or Quadrant 3.
${\sin}^{2} A = 1 - {\cos}^{2} A = 1 - \frac{16}{25} = \frac{9}{25}$
$\sin A = \pm \frac{3}{5}$
$\sin 2 A = 2 \sin A . \cos A = 2 \left(\pm \frac{3}{5}\right) \left(- \frac{4}{5}\right) = \pm \frac{24}{25}$
${\cos}^{2} A = 1 - {\sin}^{2} 2 A = 1 - \left(\frac{576}{625}\right) = \frac{49}{625}$
$\cos 2 A = \pm \frac{7}{25}$
$\tan 2 A = \frac{\sin 2 A}{\cos 2 A} = \left(\pm \frac{24}{25}\right) \left(\pm \frac{25}{7}\right) = \pm \frac{24}{7}$
If A lies in Quadrant 2, then, 2A lies in Quadrant 3, and tan 2A is positive
If A lies in Q. 3, then, 2A lies in Q. 1, then, tan 2A is positive.
Finally
$\tan 2 A = \frac{24}{7}$