# Double Angle Formulas: Finding $\tan 2\theta$

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?

• $(5/12)^2$ is not $25/12$. It is $25/144$. – Nick Nov 22 '18 at 0:12
• Erm, that's probably my issue then. – LuminousNutria Nov 22 '18 at 0:12

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 $$\tan 2\theta = \sin 2\theta/ \cos 2 \theta = 120/119$$.

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, $$\tan 2\theta = \frac{120}{119}$$