If sin x = -2/3 and x is in the third quadrant, how do you find the value of sin 2x?

Nov 6, 2016

$\sin 2 x = \frac{4 \sqrt{5}}{9}$

Explanation:

$\sin 2 x$ is equivalent to $2 \sin x \cos x$.

We must hence find cosine of $x$ to evaluate.

Letting $x$ be the adjacent side, $y$ be the opposite side and $r$ be the hypotenuse, by pythagorean theorem we have:

${x}^{2} + {y}^{2} = {r}^{2}$

${x}^{2} + {\left(- 2\right)}^{2} = {3}^{2}$

${x}^{2} = 5$

$x = \pm \sqrt{5}$

Since we're in quadrant $3$, cosine is negative. It is impossible to have a negative hypotenuse, so $x$ measures $\sqrt{5}$ units.

Hence, $\cos x = - \frac{\sqrt{5}}{3}$

We can now evaluate.

$\sin \left(2 x\right) = 2 \sin x \cos x = 2 \left(- \frac{2}{3}\right) \left(- \frac{\sqrt{5}}{3}\right) = \frac{4 \sqrt{5}}{9}$

Hopefully this helps!