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 ?

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.
