6
$\begingroup$

I would like to solve:

$ x +y+z=\frac{11\pi}{6} $

$ \sin(x)+\sin(y)+\sin(z)= \frac{\sqrt{3}}{2} $

$ \cos(x)+\cos(y)+\cos(z)=\frac{1}{2} $

After eliminating $ z $ I get:

$ 2\sin(x)+2\sin(y)-\cos(x+y)-\sqrt{3}\sin(x+y)=\sqrt{3}\tag{1}$

$ 2\cos(x)+2\cos(y)+\sqrt{3}\cos(x+y)-\sin(x+y)=1\tag{2}$

Also: $\sqrt{3}\times(1)+(2)$ gives $ \cos(x-\frac{\pi}{3})+\cos(y-\frac{\pi}{3})-\sin(x+y)=1 $...

My attempt:

$ \sin(x)+\sin(y)+\sin(z)=\sin(x+y+z+\frac{\pi}{2}) $

$ \sin(x)+\sin(y)+\sin(z)-\sin(x+y+z+\frac{\pi}{2})=0 $

$(1)$: $ \sin(\frac{x+y}{2})\cos(\frac{x-y}{2})-\sin(\frac{x+y}{2}+\frac{\pi}{4})\cos(\frac{x+y+2z}{2}+\frac{\pi}{4})=0 $

$ \cos(x)+\cos(y)+\cos(z)-\cos(x+y+z+\frac{\pi}{2})=0 $

$ (2) $: $ \cos(\frac{x+y}{2})\cos(\frac{x-y}{2})+\sin(\frac{x+y+2z}{2}+\frac{\pi}{4})\sin(\frac{x+y}{2}+\frac{\pi}{4})=0 $

$(1)+(2)$: $ \cos(\frac{x-y}{2})\cos(\frac{x+y}{2}-\frac{\pi}{4})+\sin(\frac{x+y}{2}+\frac{\pi}{4})\sin(\frac{x+y+2z}{2})=0 $

$ \cos(\frac{x+y}{2}-\frac{\pi}{4})(\cos(\frac{x-y}{2})+\sin(\frac{x+y+2z}{2}))=0 $

$ \cos(\frac{x+y}{2}-\frac{\pi}{4})(\sin(\frac{x-y}{2}+\frac{\pi}{2})+\sin(\frac{x+y+2z}{2}))=0 $

$ \cos(\frac{x+y}{2}-\frac{\pi}{4})\sin(\frac{x+z}{2}+\frac{\pi}{4})\cos(\frac{y+z}{2}-\frac{\pi}{4})=0 $

$ x+y=-\frac{\pi}{2} (\mod2\pi) $

$ x+z=-\frac{\pi}{2} (\mod2\pi) $

$ y+z=-\frac{\pi}{2} (\mod2\pi) $

Using only these equations to determine $ x,y,z $:

$ x=y=z=-\frac{\pi}{4} (\mod\pi) $ and the system is not satisfied.

Using $ x+y+z=\frac{11\pi}{6} $ to determine one of the three quantities:

$x=y=-\frac{\pi}{4} (\mod\pi) $, $ z=\frac{\pi}{3} (\mod\pi) $ or $x=z=-\frac{\pi}{4} (\mod\pi) $, $ y=\frac{\pi}{3} (\mod\pi) $ or $y=z=-\frac{\pi}{4} (\mod\pi) $, $ x=\frac{\pi}{3} (\mod\pi) $

  • 1
    You could add the Pythagorean relation, so that you have four equations in the four unknowns $\sin\,x,\sin\,y,\cos\,x,\cos\,y$...2011-08-31
  • 0
    ...anyway, the triple $(\pi/4,\pi/3,5\pi/4)$ should be the solution you're obtaining.2011-08-31
  • 4
    The right-hand sides of your last two equations are, respectively, $\sin(\pi/3)$ and $\cos(\pi/3)$. If one of our angles *happens* to be $\pi/3$, then we're in luck, as it alone on the left-hand sides balances the right-hand values. We now must make the sines, and the cosines, of the other two angles cancel; this requires that those angles differ by $\pi$. Knowing (from the first equation) that their sum is $3\pi/2$, @J.M.'s solution follows. (This opportunistic approach doesn't rule out other possible solutions, but I'll withhold a rigorous answer until I'm sure this isn't homework.)2011-08-31

1 Answers 1