MHB Solution to Trigonometric System

AI Thread Summary
The discussion focuses on finding real values of x, y, and z within the interval [0, π/2] that satisfy the trigonometric equations sin(x)cos(y) = sin(z) and cos(x)sin(y) = cos(z). Participants evaluate different approaches to solving the system, with one user acknowledging another's method while pointing out an oversight in identifying all possible solutions. The conversation highlights the collaborative nature of problem-solving in mathematics, as users share insights and corrections. Ultimately, the aim is to identify all valid pairs of solutions to the given equations. The thread emphasizes the importance of thoroughness in mathematical problem-solving.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Find all reals $$x,\,y,\,z \in \left[0,\,\frac{\pi}{2}\right]$$ that satisfying the system below:

$\sin x \cos y=\sin z\\\cos x \sin y=\cos z$
 
Mathematics news on Phys.org
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.
 
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 $x=z = \frac{\pi}{2}, y= 0$

- - - Updated - - -

Theia said:
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
$\sin \,y = \cos\, x = 0$
 
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:

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:

$$\left(x,\,y,\,z\right)=\left(0,\,\frac{\pi}{2},\,0\right),\,\left(\frac{\pi}{2},\,0,\,\frac{\pi}{2}\right)$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top