Precalculus (6th Edition)

$\displaystyle \cos 2\theta=\frac{119}{169}$ $\sin 2\displaystyle \theta=-\frac{120}{169}$
First, using the Pythagorean Identity $\sin^{2}\theta+\cos^{2}\theta=1,$with $\sin\theta > 0,$ $\displaystyle \sin\theta=+\sqrt{1-(-\frac{12}{13})^{2}}=\sqrt{1-\frac{144}{169}}$ $=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}$ We now use the Double-Angle Identities: $\displaystyle \cos 2\theta=\cos^{2}\theta-\sin^{2}\theta=\frac{144}{169}-\frac{25}{169}=\frac{119}{169}$ $\sin 2\displaystyle \theta=2\sin\theta\cos\theta=2\cdot\frac{-12}{13}\cdot\frac{5}{13}=-\frac{120}{169}$