Double Angle Formulas: Finding $\tan 2\theta$
Andrew Henderson 
I am trying to find $\tan 2\theta$ where $sin \theta = \frac{5}{13}$ and $\theta$ is in Quadrant One.
According to my textbook, $\tan 2\theta = \frac{120}{119}$, but I get $\frac{-10}{13}$ instead.
The Identity I am using:
$$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^{2}\theta}$$
My Process:
Since $y = 5\;$ and $r = 13,\; x = 12.$
Apply Tangent Double Angle Formula:$$\frac{2(\frac{5}{12})}{1 - (\frac{5}{12})^2}$$
$$\frac{\frac{10}{12}}{1 - \frac{25}{12}} \cdot \frac{12}{12}$$
$$\frac{10}{12-25}$$
$$\frac{-10}{13}$$
What am I doing wrong?
$\endgroup$ 22 Answers
$\begingroup$Alternatively:
$\sin\theta = 5/13$, implies $\cos\theta = \sqrt{1-(5/13)^2} = 12/13$.
$\sin 2\theta = 2\sin \theta \cos \theta = 120/169$.
$\cos 2\theta = 2 \cos^2 \theta-1 = 1 - 2 \sin^2 \theta = 119/169$.
This gives $\tan 2\theta = \sin 2\theta/ \cos 2 \theta = 120/119$.
$\endgroup$ $\begingroup$I made a mistake solving the problem. $(\frac{5}{12})^2 \neq \frac{25}{12}$.
Actually, $(\frac{5}{12})^2 = \frac{25}{144}$.
Taking that into account:
$$\frac{\frac{10}{12}}{1 - \frac{25}{144}} \cdot \frac{12}{12}$$
$$\frac{10}{12-\frac{300}{144}}$$
$$\frac{10}{12 - \frac{25}{12}} \cdot \frac{12}{12}$$
$$\frac{120}{144-25}$$
$$\frac{120}{119}$$
Therefore, $\tan 2\theta = \frac{120}{119}$
$\endgroup$
