# How do you use an appropriate Half-Angle Formula to find the exact value of the expression sin ((13pi)/12)?

Jun 29, 2015

Use $\sin \left(\frac{1}{2} \left(\frac{13 \pi}{6}\right)\right) = - \sqrt{\frac{1 - \cos \left(\frac{13 \pi}{6}\right)}{2}}$ to get $- \frac{\sqrt{2 - \sqrt{3}}}{2}$. (Which can be written in other forms.)

#### Explanation:

Because we want a sine, the appropriate Half-Angle Formula is:

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

We note that $\frac{13 \pi}{12} > \pi = \frac{12 \pi}{12}$ but $< \frac{3 \pi}{2} = \frac{18 \pi}{12}$
So it is in the third quadrant and has negative sine.

To find $x$, we ask oursolves "Ok, so $\frac{13 \pi}{12}$ is half of what?"

We could solve $\frac{1}{2} x = \frac{13 \pi}{12}$, but it is pretty clear that $\frac{1}{2} \cdot \frac{13 \pi}{6} = \frac{13 \pi}{12}$, so $x = \frac{13 \pi}{6}$
(Just double the number we're interested in.)

Finally, we note that $\cos \left(\frac{13 \pi}{6}\right) = \frac{\sqrt{3}}{2}$

So we get:
$\sin \left(\frac{1}{2} \left(\frac{13 \pi}{6}\right)\right) = - \sqrt{\frac{1 - \cos \left(\frac{13 \pi}{6}\right)}{2}}$

$= - \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$

$= - \sqrt{\frac{2 - \sqrt{3}}{4}}$

$= - \frac{\sqrt{2 - \sqrt{3}}}{2}$