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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 ?

Michael Hardy
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math110
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    Your initial condition is equivalent to $\sin z \geq \sin y \geq \sin x$... – Benjamin Dickman Jun 25 '13 at 02:09
  • @B.D : No, it's not. THe conditions that $\sin z-\sin y\ge0$ and $\sin y-\sin x\ge 0$, taken together, are equivalent to the two inequalities that you write. But you seem to have missed the "$=$". That's additional information, going substantially beyond those two inequalities. – Michael Hardy Jun 25 '13 at 02:23
  • Right, I should have said "implies"; thanks. – Benjamin Dickman Jun 25 '13 at 02:25
  • The result your trying to get should not depend on which parametrization of the circle is used, so I wouldn't bother with expressions like $\cos(x-z)$. Say you have $c-b=b-a\ge0$, and $a,b,c$ are the three sines. Then the three cosines are $\sqrt{1-a^2}>\sqrt{1-b^2}>\sqrt{1-c^2}$. Look at the differences between consecutive pairs of those and see if rationalizing the numerator gets you somewhere. – Michael Hardy Jun 25 '13 at 02:35
  • oh,I known,$\cos{x-z}=1$ then $x=z+2k\pi$ – math110 Jun 25 '13 at 07:36

3 Answers3

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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.

Thomas Belulovich
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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.

vadim123
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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.

dfeuer
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