# Trigonometric system

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

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You could add the Pythagorean relation, so that you have four equations in the four unknowns $\sin\,x,\sin\,y,\cos\,x,\cos\,y$... – J. M. Aug 31 '11 at 8:13
...anyway, the triple $(\pi/4,\pi/3,5\pi/4)$ should be the solution you're obtaining. – J. M. Aug 31 '11 at 9:59
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.) – Blue Aug 31 '11 at 13:22

Multiply the 2nd equation by $i$ and add the 3rd equation to get $$e^{ix}+e^{iy}+e^{iz}=e^{i\pi/3}$$ Now you can argue geometrically that for three numbers of modulus 1 to sum to a number of modulus 1, one of the summands must be the negative of another summand (and so the third summand must equal the lone number on the other side). From there, it's easy.
If you don't like geometry, you can rewrite as $$e^{ix}+e^{iy}=e^{i\pi/3}-e^{iz}$$ then express the left side as $e^{is}\cos t$ and the right side as $e^{iu}\sin v$ for appropriate $s,t,u,v$, and that should give you enough relations among $x,y,z$ to solve the problem.