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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

Csaba Daniel Farkaš
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3 Answers3

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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 $n$

Anas A. Ibrahim
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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

Asv
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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, $y=2m\pi-\dfrac\pi2+x$

lab bhattacharjee
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