This... is... TRIGONOMETRY!

Geometry Level 5


{tanxtany=sinzcosxcosy+3tanytanz=sinxcosycosz5tanztanx=sinycoszcosx3,\begin{cases} \tan { x } \tan { y } = \frac { \sin { z } }{\cos{x}\cos{y} } +3 \\ \tan{y}\tan{z}=\frac {\sin{x}}{\cos{y}\cos{z}}-5 \\ \tan{z}\tan{x}=\frac {\sin{y}}{\cos{z}\cos{x}}-3 \end{cases},


x=a+bm;x=a+bm; g+hmg+hm,

y=c+dn;y=c+dn; i+jni+jn,

z=e+f(2pmn);z=e+f(2p-m-n); k+l(2pmn+1)k+l(2p-m-n+1).

where a,b,c,d,e,f,g,h,i,j,k,a,b,c,d,e,f,g,h,i,j,k, and ll are constants, and n,m,pZn,m,p\in \mathbb{Z} ,


abcdefghijkl+π\left| \left\lfloor a\cdot b\cdot c\cdot d\cdot e\cdot f\cdot g\cdot h\cdot i\cdot j\cdot k\cdot l+\pi \right\rfloor \right| .


  1. m,m, n,n, and p,p, represent the multiplicities of periods of solutions. Ex: If sin(x)=0\sin{(x)}=0, then x=πn,nZx=\pi n, n\in \mathbb{Z}.

  2. m,m, n,n, and p,p, encompass all integers. So 2p2p would go: ...,2,0,2,4,......,-2,0,2,4,....

  3. Notice that in the presented form of solutions, constants like b,b, d,d, and ff are uniquely determined. While f(2p)=(2f)p,f*(2p)=(2f)*p, as (2f)p(2f)*p ff would not factor out nicely with mm and nn... etc.

  4. If any constant turns out to be 0 (one of them does), please treat it as 1. I've just noticed this, so if anyone entered 3, you're right. Sorry about that!

  5. Please enter the solution expansion points based on the domain of the principal value of the relevant trigonometric function (defined by the range of its inverse function). For example, for sin(x)=1\sin{(x)}=-1, x=π2+2πnx=-\frac{\pi}{2}+2\pi n (the principal range of sin1(x)\sin^{-1}{(x)} is π2xπ2-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}).

[HINT: all constants π\leq \pi.]


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