Given that $x$ and $y$ are elements of $[0, \frac{\pi}{2}]$ such that $\sin x = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\cos y = \frac{\sqrt{3}}{2}$. We need to: 1. Verify that $(\frac{\sqrt{6} + \sqrt{2}}{4})^2 = \frac{2 - \sqrt{3}}{4}$.

1. Problem Description

Given that $x$ and $y$ are elements of $[0, \frac{\pi}{2}]$ such that $\sin x = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\cos y = \frac{\sqrt{3}}{2}$.

We need to:

1. Verify that $(\frac{\sqrt{6} + \sqrt{2}}{4})^2 = \frac{2 - \sqrt{3}}{4}$.

2. Calculate $\cos x$ and $\sin y$. Determine the value of $y$.

3. Calculate $\cos(x+y)$ and $\sin(x+y)$.

4. Calculate $\cos(x-y)$ and $\sin(x-y)$. Deduce the value of $x$.

2. Solution Steps

1. Verification:

$(\frac{\sqrt{6} + \sqrt{2}}{4})^2 = \frac{(\sqrt{6} + \sqrt{2})^2}{4^2} = \frac{6 + 2\sqrt{12} + 2}{16} = \frac{8 + 2\sqrt{4 \cdot 3}}{16} = \frac{8 + 4\sqrt{3}}{16} = \frac{2 + \sqrt{3}}{4}$.

However, we need to verify $(\frac{\sqrt{6} + \sqrt{2}}{4})^2 = \frac{2 - \sqrt{3}}{4}$. We made a mistake here. It seems the problem statement has a typo. It should be $(\frac{\sqrt{6} - \sqrt{2}}{4})^2 = \frac{2-\sqrt{3}}{4}$.

$(\frac{\sqrt{6} - \sqrt{2}}{4})^2 = \frac{(\sqrt{6} - \sqrt{2})^2}{16} = \frac{6 - 2\sqrt{12} + 2}{16} = \frac{8 - 4\sqrt{3}}{16} = \frac{2 - \sqrt{3}}{4}$. So the given equation needs correction. We proceed with $\sin x = \frac{\sqrt{6} + \sqrt{2}}{4}$.

2. Calculate $\cos x$ and $\sin y$:

Since $x \in [0, \frac{\pi}{2}]$, $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (\frac{\sqrt{6} + \sqrt{2}}{4})^2} = \sqrt{1 - \frac{2 + \sqrt{3}}{4}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$.

Since $\cos y = \frac{\sqrt{3}}{2}$ and $y \in [0, \frac{\pi}{2}]$, $y = \frac{\pi}{6}$ (or 30 degrees).

Then $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - (\frac{\sqrt{3}}{2})^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

3. Calculate $\cos(x+y)$ and $\sin(x+y)$:

$\cos(x+y) = \cos x \cos y - \sin x \sin y = (\frac{\sqrt{2 - \sqrt{3}}}{2})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{6} + \sqrt{2}}{4})(\frac{1}{2}) = \frac{\sqrt{3}\sqrt{2 - \sqrt{3}}}{4} - \frac{\sqrt{6} + \sqrt{2}}{8}$.

$\sin(x+y) = \sin x \cos y + \cos x \sin y = (\frac{\sqrt{6} + \sqrt{2}}{4})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2 - \sqrt{3}}}{2})(\frac{1}{2}) = \frac{\sqrt{18} + \sqrt{6}}{8} + \frac{\sqrt{2 - \sqrt{3}}}{4} = \frac{3\sqrt{2} + \sqrt{6}}{8} + \frac{\sqrt{2 - \sqrt{3}}}{4}$.

4. Calculate $\cos(x-y)$ and $\sin(x-y)$:

$\cos(x-y) = \cos x \cos y + \sin x \sin y = (\frac{\sqrt{2 - \sqrt{3}}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{6} + \sqrt{2}}{4})(\frac{1}{2}) = \frac{\sqrt{3}\sqrt{2 - \sqrt{3}}}{4} + \frac{\sqrt{6} + \sqrt{2}}{8}$.

$\sin(x-y) = \sin x \cos y - \cos x \sin y = (\frac{\sqrt{6} + \sqrt{2}}{4})(\frac{\sqrt{3}}{2}) - (\frac{\sqrt{2 - \sqrt{3}}}{2})(\frac{1}{2}) = \frac{\sqrt{18} + \sqrt{6}}{8} - \frac{\sqrt{2 - \sqrt{3}}}{4} = \frac{3\sqrt{2} + \sqrt{6}}{8} - \frac{\sqrt{2 - \sqrt{3}}}{4}$.

We know that $\sin x = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{3} + 1)}{4} = \frac{\sqrt{2}}{2} \frac{\sqrt{3} + 1}{2}$. We also know that $\sin \frac{5\pi}{12} = \sin (75^{\circ}) = \sin (45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ} = \frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}$. Thus $x = \frac{5\pi}{12}$.

3. Final Answer

1. $(\frac{\sqrt{6} + \sqrt{2}}{4})^2 = \frac{2 + \sqrt{3}}{4}$

2. $\cos x = \frac{\sqrt{2 - \sqrt{3}}}{2}$, $\sin y = \frac{1}{2}$, $y = \frac{\pi}{6}$

3. $\cos(x+y) = \frac{\sqrt{3}\sqrt{2 - \sqrt{3}}}{4} - \frac{\sqrt{6} + \sqrt{2}}{8}$, $\sin(x+y) = \frac{3\sqrt{2} + \sqrt{6}}{8} + \frac{\sqrt{2 - \sqrt{3}}}{4}$

4. $\cos(x-y) = \frac{\sqrt{3}\sqrt{2 - \sqrt{3}}}{4} + \frac{\sqrt{6} + \sqrt{2}}{8}$, $\sin(x-y) = \frac{3\sqrt{2} + \sqrt{6}}{8} - \frac{\sqrt{2 - \sqrt{3}}}{4}$, $x = \frac{5\pi}{12}$