# If $x \leq y \leq z$, and both sides of the equation are defined, then $\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.$ [duplicate]

Show that if $x$, $y$, and $z$ are consecutive terms of an arithmetic sequence, with $x \leq y \leq z$, and both sides of the equation are defined, then $$\frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z} = \tan y.$$

I have no idea how to even start this problem, I'm stuck. Solutions are greatly appreciated.

## marked as duplicate by user219577, Willie Wong, Community♦Nov 8 '16 at 21:39

You can take the terms of the arithmetic progression as $x=a-h,y=a,z=a+h$ then you have $$\sin(a-h)=\sin a\cosh-\cos a\sin h\\\sin(a+h)=\sin a\cos h+\cos a\sin h\\\cos(a-h)=\cos a\cos h+\sin a\sin h\\\cos(a+h)=\cos a\cos h-\sin a\sin h$$ It follows $$\frac{\sin(a-h)+\sin a+\sin(a+h)}{\cos(a-h)+\cos a+\cos(a+h)}=\frac{\sin a(2\cos h+1)}{\cos a (2\cos h+1)}=\frac{\sin a}{\cos a}= \tan a$$
• You hsd (and still have) $\cosh$ instead of $\cos h$ – steven gregory Nov 4 '16 at 19:04
Let $x=y-d$ and $z=y+d$ with $d>0$. Then \begin{align*} \frac{\sin x + \sin y + \sin z }{\cos x + \cos y + \cos z}& = \frac{\sin (y-d) + \sin y + \sin (y+d) }{\cos (y-d) + \cos y + \cos (y+d)}\\ & = \frac{2\sin y \cos d + \sin y}{2 \cos y \cos d + \cos y}\\ & = \tan y. \end{align*}