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.
 A: If $x$, $y$, and $z$ are an arithmetic sequence, then for some $c$, $x = y-c$ and $z = y+c$. Substituting, that's
$\frac{\sin (y-c) + \sin(y)+\sin(y+c)}{\cos(y-c)+\cos(y)+\cos(y+c)}$
Then, apply the sum/difference identities and the answer should come quickly
A: Hint: write $x=y-d$ and $z=y+d$. The numerator can be written
$$
\sin(y-d)+\sin y+\sin(y+d)=
\sin y\cos d+\sin y+\sin y\cos d=\sin y(1+2\cos d)
$$
while the denominator is
$$
\cos(y-d)+\cos y+\cos(y+d)=…
$$
A: Write $x=y-d$, $z=y+d$. Then expand $\sin(x-d)$, $\sin(x+d)$, etc.
A: 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*}
