# If cosx=1/2, how do you find sin2x?

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

#### Explanation:

Let's first convert $\sin 2 x$ into something a bit more friendly to work with. There is an identity that states:

$\sin 2 x = 2 \sin x \cos x$

For this next bit, I'm going to focus on Quadrant 1 so that we can figure out the lengths of the sides. Once we have that, then we'll worry about other quadrants.

We're given $\cos x$ so we only need to find $\sin x$. We can do that by realizing that $\cos x = \text{adj"/"hyp} = \frac{1}{2}$ and from the Pythagorean Theorem:

${a}^{2} + {b}^{2} = {c}^{2}$

${1}^{2} + {b}^{2} = {2}^{2}$

$1 + {b}^{2} = 4$

${b}^{2} = 3$

$b = \sqrt{3}$

(We could also have known this by remembering about the 30, 60, 90 triangle).

We can now see that $\sin x = \frac{\sqrt{3}}{2}$.

Now let's work through the quadrants.

When $\cos x$ is positive, it can be in the first or fourth quadrant. Now, in the first quadrant, $\sin x$ is also positive, but in the fourth quadrant, $\sin x$ is negative. This means that when $\cos x = \frac{1}{2}$, $\sin x = \pm \frac{\sqrt{3}}{2}$.

We can then substitute in:

$\sin 2 x = 2 \left(\pm \frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right) = \pm \frac{\sqrt{3}}{2}$