The problem asks us to calculate $cis(72^\circ)$. The notation $cis(\theta)$ is equivalent to $cos(\theta) + i \cdot sin(\theta)$. Therefore, we need to calculate $cos(72^\circ) + i \cdot sin(72^\circ)$.

1. Problem Description

The problem asks us to calculate $cis(72^\circ)$. The notation $cis(\theta)$ is equivalent to $cos(\theta) + i \cdot sin(\theta)$. Therefore, we need to calculate $cos(72^\circ) + i \cdot sin(72^\circ)$.

2. Solution Steps

We need to determine $cos(72^\circ)$ and $sin(72^\circ)$. Recall that $72^\circ = \frac{2\pi}{5}$ radians.

Let $\theta = 72^\circ$. Then $5\theta = 360^\circ$. Thus $2\theta = 360^\circ - 3\theta$. Taking sine of both sides, we have

$sin(2\theta) = sin(360^\circ - 3\theta) = -sin(3\theta)$.

Therefore, $2sin(\theta)cos(\theta) = -3sin(\theta) + 4sin^3(\theta)$.

Since $sin(\theta) \ne 0$, we can divide both sides by $sin(\theta)$ to get

$2cos(\theta) = -3 + 4sin^2(\theta) = -3 + 4(1-cos^2(\theta)) = -3 + 4 - 4cos^2(\theta) = 1 - 4cos^2(\theta)$.

So $4cos^2(\theta) + 2cos(\theta) - 1 = 0$.

Let $x = cos(\theta)$. Then $4x^2 + 2x - 1 = 0$.

$x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-2 \pm 2\sqrt{5}}{8} = \frac{-1 \pm \sqrt{5}}{4}$.

Since $cos(72^\circ) > 0$, we have $cos(72^\circ) = \frac{-1+\sqrt{5}}{4}$.

Also, $sin^2(72^\circ) = 1 - cos^2(72^\circ) = 1 - (\frac{-1+\sqrt{5}}{4})^2 = 1 - \frac{1 - 2\sqrt{5}+5}{16} = 1 - \frac{6 - 2\sqrt{5}}{16} = \frac{16 - 6 + 2\sqrt{5}}{16} = \frac{10+2\sqrt{5}}{16} = \frac{5+\sqrt{5}}{8}$.

$sin(72^\circ) = \sqrt{\frac{5+\sqrt{5}}{8}} = \frac{\sqrt{10+2\sqrt{5}}}{4}$.

Therefore, $cis(72^\circ) = cos(72^\circ) + i \cdot sin(72^\circ) = \frac{-1+\sqrt{5}}{4} + i\frac{\sqrt{10+2\sqrt{5}}}{4}$.

3. Final Answer

$\frac{-1+\sqrt{5}}{4} + i\frac{\sqrt{10+2\sqrt{5}}}{4}$