# How prove this $\cos{x}+\cos{y}+\cos{z}=1$

Question:

let $x,y,z\in R$ and such $x+y+z=\pi$,and such $$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$ show that $$\cos{x}+\cos{y}+\cos{z}=1$$

My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$ then $$a+b+c=\pi$$ and $$\tan{\dfrac{a}{4}}+\tan{\dfrac{b}{4}}+\tan{\dfrac{c}{4}}=1$$ we only prove $$\cos{\dfrac{b+c}{2}}+\cos{\dfrac{a+c}{2}}+\cos{\dfrac{a+b}{2}}=1$$ Use $$\cos{\dfrac{\pi-x}{2}}=\sin{\dfrac{x}{2}}$$ $$\Longleftrightarrow \sin{\dfrac{a}{2}}+\sin{\dfrac{b}{2}}+\sin{\dfrac{c}{2}}=1$$ let $$\tan{\dfrac{a}{4}}=A,\tan{\dfrac{b}{4}}=B,\tan{\dfrac{\pi}{4}}=C$$ then $$A+B+C=1$$ and use $$\sin{2x}=\dfrac{2\tan{x}}{1+\tan^2{x}}$$ so we only prove $$\dfrac{2A}{1+A^2}+\dfrac{2B}{1+B^2}+\dfrac{2C}{1+C^2}=1$$

other idea:let $$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$ then we have $$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$ we only prove $$\cos{(2(b+c)}+\cos{2(a+c)}+\cos{2(a+b)}=\sin{(2a)}+\sin{(2b)}+\sin{(2c)}=1$$ then I fell very ugly, can you some can help?

Thank you very much!

• What were the reasons invoked for temporarily suspending your account? – Lucian Apr 11 '14 at 15:21
• maybe I do somewrong thing.sorry – math110 Apr 11 '14 at 15:41
• @math110 The problem with your attempt is considering $\cos\dfrac{b+c}{2}$ as $\sin\dfrac{a}{2}$ which is not always true since $x,y,z\in\mathbb{R}$. I tried to help you using geometry approach by considering $x,y,z$ as the angles in the triangle but then I realized it is impossible because $x,y,z\notin\mathbb{Z}_+$. – Tunk-Fey Apr 11 '14 at 17:06
• why is not true? – math110 Apr 11 '14 at 17:10
• @Tunk-Fey The relation $\cos(\pi/2-x)=\sin x$ holds for any value of $x$; it doesn't matter at all whether $x$ is the angle of a triangle. – egreg Apr 11 '14 at 18:16

Looking down at the positive octant , ( arrow tips are coordinate axes).

$x+y+z=\pi$ ( the cyan colored plane. )

$\cos{x}+\cos{y}+\cos{z}=1$ , ( the pink colored area. )

$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$ , (the light gray colored area. )

I can see three solutions where the gray central area, surrounded by pink triangle meets the cyan plane.

Just a picture!

• This picture is very beautiful! – math110 Apr 12 '14 at 2:00
• Indeed, but even such a proof without words IMHO needs to be somewhat more verbose to be convincing .. – Han de Bruijn Apr 12 '14 at 12:31
• The equation: $\cos{x}+\cos{y}+\cos{z}=1$ is an infinite lattice of octahedra-like surfaces in space. Along the plane: $x + y + z = \pi$ I would concentrate my attention on the lines: $(t,\pi - t, 0)$, $(t,0,\pi - t)$ , and $(0,t,\pi - t)$ the conditions of the theoerm seem to hold. ( I am limiting myself to a description of the diagram, there may be more to prove, for example, translation along these three lines). – Alan Apr 12 '14 at 15:45

An interesting view of the surface $\cos x + \cos y + \cos z = 1$ Three lines ( in red ) are added for clarity. We can see that even along the plane $x + y + z = \pi$ we must add a network of lines to encompass the simple periodicity condition.

Now,I have solution this problem:let $x,y,z\in R$ and such $x+y+z=\pi$,and $$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$ show that $$\cos{x}+\cos{y}+\cos{z}=1$$ $$\dfrac{y+z-x}{4}=a,\dfrac{x+z-y}{4}=b,\dfrac{x+y-z}{4}=c$$ we have $$a+b+c=\dfrac{\pi}{4},\tan{a}+\tan{b}+\tan{c}=1$$ we only prove following $$\cos{(2(b+c)}+\cos{2(a+c)}+\cos{2(a+b)}=\sin{(2a)}+\sin{(2b)}+\sin{(2c)}=1$$ since $$\tan{a}+\tan{b}+\tan{(\dfrac{\pi}{4}-a-b)}=1\Longrightarrow \tan{a}+\tan{b}+\dfrac{1-\tan{(a+b)}}{1+\tan{(a+b)}}=1$$ $$\tan{a}+\tan{b}=\dfrac{2\tan{(a+b)}}{1+\tan{(a+b)}}$$ $$\Longrightarrow 1=\tan{a}+\tan{b}-\tan{a}\tan{b}$$ then $$\sin{(a+b)}=\cos{(a-b)}$$ other hand we have \begin{align*}\sin{(2x)}+\sin{(2y)}+\sin{(2z)}&=2\sin{(x+y)}\cos{(x-y)}+1-2\sin^2{(x+y)}\\ &=2\sin{(x+y)}[\cos{(x-y)}-\sin{(x+y)}]+1\\ &=1 \end{align*}