## anonymous 4 years ago Fool's problem of the day, If $$\cos x +\cos y +\cos z =0$$ and $$\sin x +\sin y +\sin z =0$$, then find the value of $$\cos(x-y) + \cos (y-z) + \cos (z-x)$$ PS: This is high school level trigonometry. Enjoy!

1. mathmate

A gem!

2. anonymous

huuuuu lol

3. nikvist

$\cos{x}+\cos{y}+\cos{z}=0\quad,\quad\sin{x}+\sin{y}+\sin{z}=0$$\cos{x}+\cos{y}=-\cos{z}\quad,\quad\sin{x}+\sin{y}=-\sin{z}$$(\cos{x}+\cos{y})^2=\cos^2{z}\quad,\quad(\sin{x}+\sin{y})^2=\sin^2{z}$$(\cos{x}+\cos{y})^2+(\sin{x}+\sin{y})^2=\cos^2{z}+\sin^2{z}=1$$2+2\cos{(x-y)}=1$$\Rightarrow\quad\cos{(x-y)}=\cos{(y-z)}=\cos{(z-x)}=-1/2$$\Rightarrow\quad\cos{(x-y)}+\cos{(y-z)}+\cos{(z-x)}=-3/2$

4. anonymous

Well done nikvist!