$\sin x = \cos y, \sin y = \cos z, \sin z = \cos x$ For real numbers $x,y,z$ solve the system of equations:
$$\begin{align} \sin x = \cos y,\\
\sin y = \cos z,\\
\sin z = \cos x\end{align}$$
Source: high school olympiads, from a collection of problems for systems of equations, no unusual tricks involved. 
So far I found that if we square two equations and use the $\sin^2 x + \cos^2 x=1$ we get $\sin^2 y + \cos^2 z=1$ which yields $\sin^2 y = \sin^2 z$. Is this correct or am I missing something? I don't know how to continue
 A: I arrived at a different result: (squaring the first two and adding)
$$\sin^2(x)+\sin^2(y)=\cos^2(y)+\cos^2(z)$$
$$\implies 1-\cos^2(x)+\sin^2(y)=\cos^2(y)+\cos^2(z)$$
Now $\cos^2(x)=\sin^2(z)$ from third equation, giving
$$1+\sin^2(y)=\cos^2(y)+(\cos^2(z)+\sin^2(z)) \iff \sin^2(y)=\cos^2(y)$$
$$\implies y= \frac{\pi}{4}+\pi n \implies x=z=\frac{\pi}{4}+\pi n$$
for some integer $n$
A: As you have written we can form the following equations-
$$\sin^2(x)+\sin^2(z)=1$$
$$\sin^2(x)+\sin^2(y)=1$$
$$\sin^2(y)+\sin^2(z)=1$$
By eliminating the $\sin^2(x)$term from the second bracket, we now get these three equations
$$\sin^2(x)+\sin^2(z)=1$$
$$\sin^2(y)-\sin^2(z)=0$$
$$\sin^2(y)+\sin^2(z)=1$$
Now by adding the 2nd and 3rd equation we get the following:
$$\sin^2(x)+\sin^2(z)=1$$
$$\sin^2(y)-\sin^2(z)=0$$
$$2\sin^2(y)=1$$
Now we get that $y=\sin^{-1}(\pm\frac{\sqrt2}{2})$
Similarly through back substitution we get that both $x$ and $z$ are also equal to $\sin^{-1}(\pm\frac{\sqrt2}{2})$
Now you can find ALL the values of $x, y$ and $z$.
*NOTE: $x, y$ and $z$ need not be equal. For example- 
$$x=\frac{3\pi}{4}$$
$$y=\frac{\pi}{4}$$
$$z=\frac{7\pi}{4}$$
are also solutions
A: Please avoid squaring whenever possible as it almost all cases introduces Extraneous Roots 
$$\cos y=\sin x=\cos\left(\dfrac\pi2-x\right)$$
$$\implies y=2m\pi\pm\left(\dfrac\pi2-x\right)$$
Taking the '+' sign, $y=2m\pi+\dfrac\pi2-x$
$$\cos z=\sin y=\cos x\implies x=2n\pi\pm z$$
$$\sin z=\cos x=\cos \left(2n\pi\pm z\right)=?$$
$$\tan z=1\implies z=?$$
Please consider the '-' sign, $y=2m\pi-\dfrac\pi2+x$
