## Trigonometry (11th Edition) Clone

$\frac{\sqrt{3}}{2}$ + $\frac{1}{2} i$ is a cube root of $i$.
If $\frac{\sqrt{3}}{2}$ + $\frac{1}{2} i$ is a cube root of $i$, $(\frac{\sqrt{3}}{2}$ + $\frac{1}{2} i)^3$ will be equal to $i$. $(\frac{\sqrt{3}}{2}$ + $\frac{1}{2} i)^3$ = $(\frac{\sqrt{3}}{2})^3$ + $3 \cdot (\frac{\sqrt{3}}{2})^2 \cdot (\frac{1}{2}i)$ + $3 \cdot (\frac{\sqrt{3}}{2}) \cdot (\frac{1}{2}i)^2 + (\frac{1}{2}i)^3$ (The cube of sum of two numbers) = $\frac{3\sqrt{3}}{8}$ + $\frac{9i}{8}$ - $\frac{3\sqrt{3}}{8}$ - $\frac{i}{8}$ = $i$ Therefore, $\frac{\sqrt{3}}{2}$ + $\frac{1}{2} i$ is a cube root of $i$.