0
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ $(5/12)^2$ is not $25/12$. It is $25/144$. $\endgroup$ – Nick Nov 22 '18 at 0:12
  • $\begingroup$ Erm, that's probably my issue then. $\endgroup$ – LuminousNutria Nov 22 '18 at 0:12
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.