# How do you find the values of sin 2theta and cos 2theta when cos theta = 12/13?

Aug 4, 2018

Below

#### Explanation:

$\theta$ can be in the first quadrant $0 \le \theta \le 90$ or the fourth quadrant $270 \le \theta \le 360$

If $\theta$ is in the first quadrant,
then
$\sin \theta = \frac{5}{13}$
$\cos \theta = \frac{12}{13}$
$\tan \theta = \frac{5}{12}$

Therefore,
$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169}$

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta = {\left(\frac{12}{13}\right)}^{2} - {\left(\frac{5}{13}\right)}^{2} = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}$

If $\theta$ is in the fourth quadrant,
then
$\sin \theta = - \frac{5}{13}$
$\cos \theta = \frac{12}{13}$
$\tan \theta = - \frac{5}{12}$

Therefore,
$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \times - \frac{5}{13} \times \frac{12}{13} = - \frac{120}{169}$

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta = {\left(\frac{12}{13}\right)}^{2} - {\left(- \frac{5}{13}\right)}^{2} = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}$

Aug 4, 2018

$\sin 2 \theta = \frac{120}{169} , \theta \in {Q}_{1} \mathmr{and} - \frac{120}{169} , \theta \in {Q}_{4}$.

$\cos 2 \theta = \frac{119}{169}$

#### Explanation:

https://socratic.org/questions/if-cos-a-5-13-how-do-you-find-sina-and-tana

As a continuation,

$\sin \theta = \frac{5}{13} , \theta \in {Q}_{1}$ and it is $- \frac{5}{13} , \theta \in {Q}_{4}$.

$\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \left(\frac{5}{13}\right) \left(\frac{12}{13}\right)$

$= \frac{120}{169} , \theta \in {Q}_{1} \mathmr{and}$

$= 2 \left(- \frac{5}{13}\right) \left(\frac{12}{13}\right) = - \frac{120}{169} , \theta \in {Q}_{4}$.

$\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta = {\left(\frac{12}{13}\right)}^{2} - {\left(\pm \frac{5}{13}\right)}^{2}$

$= \frac{119}{169}$