# How do you use the half angle identity to find sin 105?

Mar 20, 2018

$\sin {105}^{\circ} = \frac{\sqrt{2 + \sqrt{3}}}{2}$

#### Explanation:

We need to use the half angle formula:

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

In this case, we want to find $\sin \left({105}^{\circ}\right)$, so that's what we want $\sin \left(\frac{\theta}{2}\right)$ to equal. To find out what our $\theta$ is, set these to equal to each other:

$\sin \left({105}^{\circ}\right) = \sin \left(\frac{\theta}{2}\right)$

${105}^{\circ} = \frac{\theta}{2}$

${210}^{\circ} = \theta$

This is our $\theta$. Now, we can use the half angle formula:

$\textcolor{w h i t e}{=} \sin \left({105}^{\circ}\right)$

$= \sin \left({210}^{\circ} / 2\right)$

$= \pm \sqrt{\frac{1 - \cos \left({210}^{\circ}\right)}{2}}$

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

$= \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

$= \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

$= \pm \sqrt{\frac{2 + \sqrt{3}}{4}}$

$= \pm \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$

$= \pm \frac{\sqrt{2 + \sqrt{3}}}{2}$

Since ${105}^{\circ}$ is in quadrant II, we know that our answer will be positive that angle is above the $x$-axis (and we are taking the sine). Therefore:

$\sin {105}^{\circ} = \frac{\sqrt{2 + \sqrt{3}}}{2}$

We can check our answer using a calculator (be sure it is in degrees mode):

Mar 20, 2018

color(blue)(sin (105) = = +-(1/2) sqrt(2 + sqrt3)

#### Explanation:

Given $\frac{\theta}{2} = {105}^{\circ} = \frac{7 \pi}{12}$

$\theta = \frac{7 \pi}{6}$

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

$\sin \left(\frac{7 \pi}{12}\right) = \pm \sqrt{\frac{1 - \cos \left(\left(\frac{7 \pi}{12}\right) \cdot 2\right)}{2}}$

$\sin \left(\frac{7 \pi}{12}\right) = \pm \sqrt{\frac{1 - \cos \left(\frac{7 \pi}{6}\right)}{2}}$

But $\cos \left(\frac{7 \pi}{6}\right) = \cos \left(\pi + \frac{\pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right)$

$\therefore \sin \left(\frac{7 \pi}{12}\right) = \pm \sqrt{\frac{1 + \cos \left(\frac{\pi}{6}\right)}{2}}$

$\implies \pm \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

$\implies \pm \sqrt{\frac{2 + \sqrt{3}}{4}} = \pm \left(\frac{1}{2}\right) \sqrt{2 + \sqrt{3}}$