Question

How do you find the values of $\sin 2\theta$ and $\cos 2\theta$ when $\cos \theta =\frac{12}{13}$ ?

Answer 1

$\theta$ 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,
$\sin 2\theta =2\sin \theta \cos \theta =2\times \frac{5}{12}\times \frac{12}{13}=\frac{120}{169}$
$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =(\frac{12}{13}{)}^2-(\frac{5}{13}{)}^2=\frac{144}{169}-\frac{25}{169}=\frac{119}{169}$
If $\theta$ is in the fourth quadrant, then
$\sin \theta =-\frac{5}{13}$
$\cos \theta =\frac{12}{13}$
$\tan \theta =-\frac{5}{12}$
Therefore,
$\sin 2\theta =2\sin \theta \cos \theta =2\times -\frac{5}{12}\times \frac{12}{13}=-\frac{120}{169}$
$\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta =(\frac{12}{13}{)}^2-(-\frac{5}{13}{)}^2=\frac{144}{169}-\frac{25}{169}=\frac{119}{169}$