Given that $\sin \alpha = \frac{\sqrt{3}}{2}$ and $0 < \alpha < \pi$, find $\cos \alpha$, $\tan \alpha$, and $\cot \alpha$. Also, prove the identity $\sin^4 x + \cos^4 x = 1 - 3\sin^2 x \cos^2 x$. This looks like a typographical error and is actually $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$.

1. Problem Description

Given that $\sin \alpha = \frac{\sqrt{3}}{2}$ and $0 < \alpha < \pi$, find $\cos \alpha$, $\tan \alpha$, and $\cot \alpha$. Also, prove the identity $\sin^4 x + \cos^4 x = 1 - 3\sin^2 x \cos^2 x$. This looks like a typographical error and is actually $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$.

2. Solution Steps

First, we find $\cos \alpha$. Since $\sin^2 \alpha + \cos^2 \alpha = 1$, we have:

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

$\cos^2 \alpha = 1 - \left(\frac{\sqrt{3}}{2}\right)^2$

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

$\cos \alpha = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$

Since $0 < \alpha < \pi$, $\alpha$ can be in the first or second quadrant.

If $\alpha$ is in the first quadrant $(0 < \alpha < \frac{\pi}{2})$, then $\cos \alpha = \frac{1}{2}$.

If $\alpha$ is in the second quadrant $(\frac{\pi}{2} < \alpha < \pi)$, then $\cos \alpha = -\frac{1}{2}$.

Since $\sin \alpha = \frac{\sqrt{3}}{2}$, we know $\alpha = \frac{\pi}{3}$ or $\alpha = \frac{2\pi}{3}$.

If $\alpha = \frac{\pi}{3}$, then $\cos \alpha = \cos \frac{\pi}{3} = \frac{1}{2}$.

If $\alpha = \frac{2\pi}{3}$, then $\cos \alpha = \cos \frac{2\pi}{3} = -\frac{1}{2}$.

Therefore, $\cos \alpha$ can be either $\frac{1}{2}$ or $-\frac{1}{2}$.

Next, we calculate $\tan \alpha$ and $\cot \alpha$ for both cases.

Case 1: $\cos \alpha = \frac{1}{2}$

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$

$\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

Case 2: $\cos \alpha = -\frac{1}{2}$

$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -\sqrt{3}$

$\cot \alpha = \frac{1}{\tan \alpha} = \frac{1}{-\sqrt{3}} = -\frac{\sqrt{3}}{3}$

Now, let's prove the identity $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$:

$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$

Since $\sin^2 x + \cos^2 x = 1$, we have:

$\sin^4 x + \cos^4 x = 1^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$

3. Final Answer

$\cos \alpha = \pm \frac{1}{2}$

If $\cos \alpha = \frac{1}{2}$, then $\tan \alpha = \sqrt{3}$ and $\cot \alpha = \frac{\sqrt{3}}{3}$.

If $\cos \alpha = -\frac{1}{2}$, then $\tan \alpha = -\sqrt{3}$ and $\cot \alpha = -\frac{\sqrt{3}}{3}$.

The correct identity is $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$.