We are given two conditions:
First, let's simplify the ratio involving the sines:
$ \frac{\sin x}{\sin y} = \frac{1/\sqrt{2}}{1/2} = \frac{1}{\sqrt{2}} \times \frac{2}{1} = \frac{2}{\sqrt{2}} = \sqrt{2} $
This gives us the equation $\sin x = \sqrt{2} \sin y$.
Using the sum $x + y = 75$, we can express $y$ as $y = 75 - x$. Substitute this into the simplified sine equation:
$ \sin x = \sqrt{2} \sin(75 - x) $
Expand $\sin(75 - x)$ using the sine subtraction formula ($\sin(A-B) = \sin A \cos B - \cos A \sin B$):
$ \sin x = \sqrt{2} (\sin 75 \cos x - \cos 75 \sin x) $
Rearrange the terms to find $\tan x$:
$ \sin x = \sqrt{2} \sin 75 \cos x - \sqrt{2} \cos 75 \sin x $
$ \sin x (1 + \sqrt{2} \cos 75) = \sqrt{2} \sin 75 \cos x $
$ \frac{\sin x}{\cos x} = \tan x = \frac{\sqrt{2} \sin 75}{1 + \sqrt{2} \cos 75} $
We need the values for $\sin 75^\circ$ and $\cos 75^\circ$. Using angle addition formulas:
Substitute these values back into the expression for $\tan x$:
$ \tan x = \frac{\sqrt{2} \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right)}{1 + \sqrt{2} \left( \frac{\sqrt{6} - \sqrt{2}}{4} \right)} = \frac{\frac{\sqrt{12} + 2}{4}}{1 + \frac{\sqrt{12} - 2}{4}} = \frac{\frac{2\sqrt{3} + 2}{4}}{\frac{4 + 2\sqrt{3} - 2}{4}} $
$ \tan x = \frac{\frac{\sqrt{3} + 1}{2}}{\frac{2 + 2\sqrt{3}}{4}} = \frac{\frac{\sqrt{3} + 1}{2}}{\frac{1 + \sqrt{3}}{2}} = 1 $
Since $\tan x = 1$, the principal value for $x$ is $45^\circ$.
Now, find $y$ using $x + y = 75^\circ$:
$ y = 75^\circ - x = 75^\circ - 45^\circ = 30^\circ $
The ratio $x : y$ is calculated as:
$ x : y = 45^\circ : 30^\circ $
Simplify the ratio by dividing both sides by their greatest common divisor, which is $15^\circ$:
$ x : y = \frac{45}{15} : \frac{30}{15} = 3 : 2 $
The given equation can be reduced to
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