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 ranges of x and y, we can notice that \(\sin x = \cos y = t > 0\).  (Namely, both are more than zero.)  If we use \(\sin^2\theta + \cos^2\theta = 1\) and \(\sin x = t; \cos y = t \), we can have the following expressions for each:

\[ \cos x = \sqrt{1-t^2};  \sin y = \sqrt{1-t^2} \]

Now, we can rewrite \(\sin (y-x)\) by the addition formula 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 \(y = \frac{\pi}{3}\).

Well, it might be tricky, but try to stick to find out every expression with \(t\)!