# How do you use demoivre's theorem to simplify (cos((5pi)/4)+isin((5pi)/4))^10?

Sep 13, 2016

${\left\{\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right\}}^{10} = i$

#### Explanation:

According to Demoivre's theorem,

${\left\{r \left(\cos \theta + i \sin \theta\right)\right\}}^{n} = {r}^{n} \left(\cos n \theta + i \sin n \theta\right)$

Hence, ${\left\{\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right\}}^{10}$

= $\left\{\cos \left(\frac{5 \pi}{4} \times 10\right) + i \sin \left(\frac{5 \pi}{4} \times 10\right)\right\}$

= $\left\{\cos \left(\frac{50 \pi}{4}\right) + i \sin \left(\frac{50 \pi}{4}\right)\right\}$

= $\left\{\cos \left(\frac{25 \pi}{2}\right) + i \sin \left(\frac{25 \pi}{2}\right)\right\}$

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

= $\left\{\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right\}$

= $\left\{0 + i \times 1\right\}$

= $i$