1. The problem is to evaluate the expression $\sin 30^\circ + i \cos 30^\circ$ where $i$ is the imaginary unit. 2. Recall the values of sine and cosine for $30^\circ$: - $\sin 30^\circ = \frac{1}{2}$ - $\cos 30^\circ = \frac{\sqrt{3}}{2}$ 3. Substitute these values into the expression: $$\sin 30^\circ + i \cos 30^\circ = \frac{1}{2} + i \frac{\sqrt{3}}{2}$$ 4. This expression is a complex number in the form $a + bi$ where $a = \frac{1}{2}$ and $b = \frac{\sqrt{3}}{2}$. 5. We can also express this complex number in polar form using Euler's formula: $$a + bi = r(\cos \theta + i \sin \theta) = r e^{i \theta}$$ where $r = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1}(\frac{b}{a})$. 6. Calculate the magnitude $r$: $$r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$$ 7. Calculate the argument $\theta$: $$\theta = \tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \tan^{-1}(\sqrt{3}) = 60^\circ$$ 8. Therefore, the complex number can be written as: $$e^{i 60^\circ} = \cos 60^\circ + i \sin 60^\circ$$ 9. Notice that the original expression $\sin 30^\circ + i \cos 30^\circ$ equals $\frac{1}{2} + i \frac{\sqrt{3}}{2}$, which is $e^{i 60^\circ}$, but with sine and cosine swapped compared to the usual Euler form. 10. The final answer is: $$\sin 30^\circ + i \cos 30^\circ = \frac{1}{2} + i \frac{\sqrt{3}}{2}$$