Given that $\sin \alpha = \frac{1}{2}$ and $\alpha$ is in the second quadrant, find $\cos \alpha$ and $\tan \alpha$.

1. Problem Description

Given that $\sin \alpha = \frac{1}{2}$ and $\alpha$ is in the second quadrant, find $\cos \alpha$ and $\tan \alpha$.

2. Solution Steps

We know that $\sin^2 \alpha + \cos^2 \alpha = 1$.

So,

$\cos^2 \alpha = 1 - \sin^2 \alpha$

$\cos^2 \alpha = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$

$\cos \alpha = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$.

Since $\alpha$ is in the second quadrant, $\cos \alpha$ is negative.

Therefore, $\cos \alpha = - \frac{\sqrt{3}}{2}$.

Next, we want to find $\tan \alpha$. We know that

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$.

$\tan \alpha = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{2} \times \frac{-2}{\sqrt{3}} = -\frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$.

3. Final Answer

$\cos \alpha = - \frac{\sqrt{3}}{2}$

$\tan \alpha = - \frac{\sqrt{3}}{3}$