# How can you use trigonometric functions to simplify  16 e^( ( 23 pi)/12 i )  into a non-exponential complex number?

Jan 13, 2016

$z = 8 \sqrt{2 + \sqrt{3}} - 8 \sqrt{2 - \sqrt{3}} i$

#### Explanation:

${e}^{\left(23 \pi\right) \frac{i}{12}} = \cos \left(23 \frac{\pi}{12}\right) + i \sin \left(23 \frac{\pi}{12}\right)$

Since $23 \frac{\pi}{12} = 2 \pi - \frac{\pi}{12}$ so we have

${e}^{\left(23 \pi\right) \frac{i}{12}} = \cos \left(2 \pi - \frac{\pi}{12}\right) + i \sin \left(2 \pi - \frac{\pi}{12}\right)$

But since $2 \pi$ is the period, we can safely ignore it

${e}^{\left(23 \pi\right) \frac{i}{12}} = \cos \left(- \frac{\pi}{12}\right) + i \sin \left(- \frac{\pi}{12}\right)$

Remember that $\cos \left(- \theta\right) = \cos \left(\theta\right)$ and $\sin \left(- \theta\right) = - \sin \left(\theta\right)$

${e}^{\left(23 \pi\right) \frac{i}{12}} = \cos \left(\frac{\pi}{12}\right) - i \sin \left(\frac{\pi}{12}\right)$

From there, it's using half angle formulas to figure out these values

$\sin \left(\frac{u}{2}\right) = \sqrt{\frac{1 - \cos \left(u\right)}{2}}$ and $\cos \left(u\right) = \sqrt{\frac{1 + \cos \left(u\right)}{2}}$

So

$\sin \left(\frac{\pi}{12}\right) = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$

$\cos \left(\frac{\pi}{12}\right) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$

${e}^{\left(23 \pi\right) \frac{i}{12}} = \frac{\sqrt{2 + \sqrt{3}}}{2} - i \frac{\sqrt{2 - \sqrt{3}}}{2}$

And lastly, multiply by $16$ so we have

$z = 8 \sqrt{2 + \sqrt{3}} - 8 \sqrt{2 - \sqrt{3}} i$