Trigonometric functions with a parametric representation

Question:
Here is a set of simultaneous equations, $\sin x = \cos y = \sin (y-x)$. The ranges are $0 < x < \frac{\pi}{2}$ and $0 < y < \frac{\pi}{2}$. Find $x$ and $y$.

Answer:
It is easier to equate the equations with a parameter. Thus,
\[ \sin x = \cos y = \sin (y-x) = t\]
From the conditions, we can notice that $\sin x = \cos y = t > 0$.  If we use $\sin^2\theta + \cos^2\theta = 1$, we can have the following expressions:
\[ \cos x = \sqrt{1-t^2};  \sin y = \sqrt{1-t^2} \]
Now, we can rewrite $\sin (y-x)$ as follows:
\[ \sin (y-x) = \sin y\cos x - \sin x\cos y  \\
                    = \sqrt{1-t^2}\cdot\sqrt{1-t^2}-t\cdot t  \\
                   = (1-t^2) - t^2  \\
                   = 1 - 2t^2 \]
Since $\sin(y-x) = t$, we have
\[ 1 - 2t^2 = t\]
Therefore,
\[ 2t^2 + t - 1 = 0  \\
    (2t - 1)(t - 1) = 0\]
The solution is that $t=\frac{1}{2}$ because $t>0$.
This gives $\sin x = \cos y = \frac{1}{2}$. Hence we obtain $x = \frac{\pi}{6}$ and $y = \frac{\pi}{3}$.

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