prove that $\cos x,\cos y,\cos z$ don't make strictly decreasing arithmetic progression 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 ?
 A: 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.
A: Suppose by way of contradiction that $\cos x, \cos y, \cos z$ were in arithmetic progression (increasing or decreasing), i.e. $$\cos x - \cos y = \cos y - \cos z $$
Multiply the equation by $-1$ and add to $i$ times $$\sin y - \sin x = \sin z - \sin y$$ to get $$e^{iy}-e^{ix}=e^{iz}-e^{iy}$$
which rearranges to $$2e^{iy}=e^{iz}+e^{ix}$$
Now $|2e^{iy}|=2$, and $|e^{iz}+e^{ix}|\le |e^{iz}|+|e^{ix}|=2$.  Hence $e^{iz}$ and $e^{ix}$ are linearly dependent as vectors, and thus equal.  But now $2e^{iy}=2e^{ix}$, so $e^{iy}=e^{ix}$.  Hence $x=y=z \pmod{2\pi}$, so $\cos x,\cos y, \cos z$ are not strictly in arithmetic progression.
A: We have $\sin y = \sin x + p$ and $\sin z = \sin x+2p$, where $p$ is positive.
Suppose that $\cos y = \cos x - r$ and $\cos z = \cos x - 2r$ for positive $r$.
Then $\cos^2 y = 1-(\sin x+p)^2 = (\cos x - r)^2$ and  $\cos^2 z = 1-(\sin x+2p)^2 = (\cos x - 2r)^2$.
So $\cos^2y = 1-\sin^2 x-p^2-2p\sin x = \cos^2 x - 2r\cos x + r^2$ and $\cos^2 z = 1-\sin^2 x - 4p^2 - 4p\sin x = \cos^2 x -4r\cos x + 4r^2$.
Then $-p^2-2p\sin x = -2r\cos x + r^2$ and $-4p^2 - 4p\sin x=-4r\cos x+4 r^2$.
That last gets you $p^2+p\sin x = r\cos x+r^2$.
I have to run now.
