# If sintheta=1/3 and theta is in quadrant I, how do you evaluate sin2theta?

Jul 2, 2016

$\frac{4 \sqrt{2}}{9}$.

#### Explanation:

The first quadrant $\theta = {\sin}^{- 1} \left(\frac{1}{3}\right) = {19.47}^{o}$, nearly. So, $2 \theta$ is

also in the first quadrant, and so, $\sin 2 \theta > 0$.

Now, $\sin 2 \theta = 2 \sin \theta \cos \theta . = 2 \left(\frac{1}{3}\right) \left(\sqrt{1 - {\left(\frac{1}{3}\right)}^{2}}\right) = \frac{4 \sqrt{2}}{9}$.

If theta is in the 2nd quadrant as $\left({180}^{o} - \theta\right)$

for which sin is $\sin \theta = \frac{1}{3}$, and $\cos \theta < 0$.

Here, $\sin 2 \theta = - \frac{4 \sqrt{2}}{9}$.