TITLE: prove that $\cos x,\cos y,\cos z$ don't make strictly decreasing arithmetic progression
QUESTION [6 upvotes]: let $x,y,z\in R$,and such that $$\sin y-\sin x=\sin z-\sin y\ge  0 $$ show  that:
$$\cos x,\cos y,\cos z$$ don't make strictly decreasing arithmetic progression
my idea:
we have $$2\sin y=\sin x +\sin z\cdots\cdots\tag 1$$
and assume that,there exist $x,y,z$ such that
$$2\cos y=\cos x+\cos z\cdots\cdots \tag2$$
and $(1)^2+(2)^2$,we have
$$4=2+2(\sin x\sin z+\cos x\cos z)=2+2\cos(x-z)$$
then
$$\cos(x-z)=1\Longrightarrow x=z+k\pi,k\in Z$$
so
$$\cos x=(-1)^k\cos z,\sin x=(-1)^k\sin z$$
Then ?

REPLY [3 votes]: The three points $(\cos x, \sin x)$, $(\cos y, \sin y)$, and $(\cos z,\sin z)$ lie on the unit circle, and by assumption are distinct. 
The y-coordinates are given to be in arithmetic progression, and we are asked to show the $x$-coordinates are not. 
If both sets of coordinates were in arithmetic progression, the three points would be collinear. A simple geometric proof would be that a line cannot intersect a circle in three points.