# If sintheta=1/3 and theta is in quadrant I, how do you evaluate tan2theta?

Apr 4, 2017

$\tan \left(2 \theta\right) = \frac{4 \sqrt{2}}{7}$

#### Explanation:

Use the equation $\tan \left(2 \theta\right) = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta}$.

The given information tells us that $\sin \theta = \frac{1}{3}$. Since $\theta$ is in the first quadrant, we know that all its trig functions will be positive.

The formula for $\tan \left(2 \theta\right)$ uses $\tan \theta$. All we have is $\sin \theta$.

From $\sin \theta$, we can figure out $\cos \theta$ using the Pythagorean identity:

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

Using $\sin \theta = \frac{1}{3}$, we see that:

$\frac{1}{9} + {\cos}^{2} \theta = 1$

$\cos \theta = \sqrt{\frac{8}{9}} = \frac{2 \sqrt{2}}{3}$

So now we can figure out $\tan \theta$:

$\tan \theta = \sin \frac{\theta}{\cos} \theta = \frac{\frac{1}{3}}{\frac{2 \sqrt{2}}{3}} = \frac{1}{3} \left(\frac{3}{2 \sqrt{2}}\right) = \frac{1}{2 \sqrt{2}}$

Applying the formula for $\tan \left(2 \theta\right)$:

$\tan \left(2 \theta\right) = \frac{2 \tan \theta}{1 - {\tan}^{2} \theta} = \frac{2 \left(\frac{1}{2 \sqrt{2}}\right)}{1 - {\left(\frac{1}{2 \sqrt{2}}\right)}^{2}}$

Simplifying:

$\tan \left(2 \theta\right) = \frac{\frac{1}{\sqrt{2}}}{1 - \frac{1}{8}} = \frac{1}{\sqrt{2}} \left(\frac{8}{7}\right) = \frac{4 \sqrt{2}}{7}$