<s>[INST] If $x$, $y$, $z$ are in arithmetic progression, show that $\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y. $ Show that if $x, y,$ and $z$ are consecutive terms of an arithmetic sequence, and $\tan y$ is defined, then 
$$\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.
$$
I'm not sure what trig identities I would use and how to use them.  Could I get some help?  Thanks.
 [/INST] Lets define the common ratio $r$
Using the 4 Trigonometric Conversion functions, we get:
\begin{eqnarray*} \sin(y+r) &=& \sin y \cos r + \cos y \sin r \\ \sin(y-r) &=& \sin y \cos r - \cos y \sin r \\ \cos(y+r) &=& \cos y \cos r - \sin y \sin r \\ \cos(y-r) &=& \cos y \cos r + \sin y \sin r
\end{eqnarray*}Since $x = y - r$, $y = y$, and $z = y+r$, we can rewrite the original equation prove:
\begin{align*}
\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.\ &= \frac{\sin (y-r) + \sin y + \sin (y+r)}{\cos (y-r) + \cos y + \cos (y+r)} \\
&= \frac{\sin y \cos r - \cos y \sin r  + \sin y + \sin y \cos r + \cos y \sin r}{\cos y \cos r + \sin y \sin r + \cos y + \cos y \cos r - \sin y \sin r } \\
&= \frac{2 \sin y \cos r + \sin y}{2 \cos y \cos r + \cos y} \\
&= \frac{\sin y (2 \cos r + 1)}{\cos y (2 \cos r + 1)} \\
&= \frac{\sin y}{\cos y} \\
&= \boxed{\tan y}
\end{align*}
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