## Trigonometry 7th Edition

Knowing that $i^{2}=-1$, we also know that $i^{4}=(i^{2})^{2}=(-1)^{2}=1$. We can use the fact that $i^{4}=1$ to evaluate higher powers of $i$. $i^{14}=i^{12}\times i^{2}=(i^{4})^{3}\times i^{2}=(1)^{3}\times-1=1\times-1=-1$