MHB Solution to Trigonometric System

[anemone]

Find all reals $$x,\,y,\,z \in \left[0,\,\frac{\pi}{2}\right]$$ that satisfying the system below:

$\sin x \cos y=\sin z\\\cos x \sin y=\cos z$

 

[Theia]

Spoiler

I can find only one solution:

Taking a square and summing up will get

$$\begin{align*}\sin^2 z + \cos^2 z = 1 &= \sin^2x\cos^2y + \cos^2x\sin^2y \\
& = \sin^2x + \sin^2y \left( \cos^2x - \sin^2x \right) \qquad \Rightarrow \end{align*} $$

$$\sin^2y = \frac{\cos^2x}{\cos^2x - \sin^2x} \ge 1.$$

Equality takes place when $$x = 0$$ and thus $$y = \frac{\pi}{2}$$ and $$z = 0$$ is the only solution.

 

[kaliprasad]

Spoiler

We have $(\sin x \cos y)^2+(\cos x \sin y)^2 =1$\
$(\sin x \cos y)^2 <= (\sin x)^2$ and $(\sin x \cos y)^2 <= (\cos y)^2$
so on we must have
either$(\sin x \cos y) = (\sin x)$ and $(\cos x \sin y) = \cos x$ $=>\sin x = 0,\sin y = 1$ or $\cos y = 1,\cos x=0$
This gives
1) $\sin x = 0,\sin y = 1$ => $x=0, y = \frac{\pi}{2}, z= 0$
2) $\cos x = 0,\cos y = 1$ => $x=z = \frac{\pi}{2}, y= 0$

Or $(\sin x \cos y) = (\cos y )$ and $(\cos x \sin y) = \sin y$ $=>\cos x = 1,\cos y = 0$ or $\sin y = 1,\cos x=1$
THis does not gives any additional solutions
so the solution set is
3)$x=0, y = \frac{\pi}{2}, z= 0$ or $x=z = \frac{\pi}{2}, y= 0$

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Spoiler

I can find only one solution:

Taking a square and summing up will get

$$\begin{align*}\sin^2 z + \cos^2 z = 1 &= \sin^2x\cos^2y + \cos^2x\sin^2y \\
& = \sin^2x + \sin^2y \left( \cos^2x - \sin^2x \right) \qquad \Rightarrow \end{align*} $$

$$\sin^2y = \frac{\cos^2x}{\cos^2x - \sin^2x} \ge 1.$$

Equality takes place when $$x = 0$$ and thus $$y = \frac{\pi}{2}$$ and $$z = 0$$ is the only solution.

Hello Theia;
your approach is better than my approach but unfortunately you overlooked

Spoiler

$\sin \,y = \cos\, x = 0$

 

[anemone]

Hi Theia!

Albeit you missed out another pair of solution, you still deserve a pat on the back for the job well done...and thanks for participating!

My solution:

Spoiler

Adding up the two equations gives:

$\sin z+\cos z=\sin x \cos y+\cos x \sin y$

Now, by applying the Cauchy-Schwarz inequality to the RHS of equation above yields:

$\sin z+\cos z=\sin x \cos y+\cos x \sin y\le \sqrt{\sin^2 x+\cos^2 x} \sqrt{\sin^2 y+\cos^2 y}=1$

$\left(\sin z+\cos z\right)^2\le 1^2$

$\sin^2 z+\cos^2 z+2\sin z \cos z\le 1^2$

$\sin z \cos z\le 0$ (*)

But since we're told $$x,\,y,\,z \in \left[0,\,\frac{\pi}{2}\right]$$, the inequality (*) holds iff $\sin z=0$ or $\cos z=0$, which lead to two set of solutions where:

$$\left(x,\,y,\,z\right)=\left(0,\,\frac{\pi}{2},\,0\right),\,\left(\frac{\pi}{2},\,0,\,\frac{\pi}{2}\right)$$