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
Best Answer
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 $n$