Question

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!

Answer 1

A gem!

Answer 2

huuuuu lol

Answer 3

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

Answer 4

Well done nikvist!