Show that if x, y, and z are consecutive terms of an arithmetic sequence, and tan y is defined, then (sin x + sin y + sin z) / (cos x + cos y + cos z) = tan y.

So I tried letting x = y-k (since x,y,z are consecutive terms of an arithmetic sequence), then z= y+k

So we get (sin(y-k) + siny + sin(y+k)) / (cos(y-k) + cos(y) + cos(y+k)) = tan y+0k

What do I do next?

numerator:

sinycosk-cosysink+siny+sinycosk+cosysink
sinycosk+siny+sinycosk
2sinycosk+siny
siny(2cosk+1)

denominator:
cosycosk+sinysink+cosy+cosycosk-sinysink
2cosycosk+cosy
cosy(2cosk+1)

so,

siny(2cosk+1)
-------------- = tany
cosy(2cosk+1)

THAT HELPED ALOT THANKS

You're on the right track! Now, let's simplify the expression on the left side of the equation.

We can use the trigonometric identity for the sum of two sine functions:

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

Applying this identity, we can rewrite the numerator of the expression as:

sin(y - k) + sin(y) + sin(y + k) = [sin(y)cos(-k) + cos(y)sin(-k)] + sin(y) + [sin(y)cos(k) + cos(y)sin(k)]

Since cosine is an even function (cos(-x) = cos(x)) and sine is an odd function (sin(-x) = -sin(x)), we can simplify further:

= sin(y)cos(k) - cos(y)sin(k) + sin(y) + sin(y)cos(k) + cos(y)sin(k)

By rearranging the terms, we have:

= sin(y)cos(k) + sin(y) + sin(y)cos(k) - cos(y)sin(k) = 2sin(y)cos(k) + sin(y) - cos(y)sin(k)

Similarly, we can apply the trigonometric identity for the sum of two cosine functions:

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

Using this identity, we can rewrite the denominator of the expression as:

cos(y - k) + cos(y) + cos(y + k) = [cos(y)cos(-k) - sin(y)sin(-k)] + cos(y) + [cos(y)cos(k) - sin(y)sin(k)]

Again, simplifying using the properties of cosine and sine functions:

= cos(y)cos(k) + sin(y)sin(k) + cos(y) + cos(y)cos(k) - sin(y)sin(k)

By rearranging the terms, we have:

= cos(y)cos(k) + cos(y) + cos(y)cos(k) + sin(y)sin(k) = 2cos(y)cos(k) + cos(y) + sin(y)sin(k)

Now we can substitute these simplified expressions back into the original equation:

(sin(y)cos(k) + sin(y) + sin(y)cos(k) - cos(y)sin(k)) / (2cos(y)cos(k) + cos(y) + sin(y)sin(k))

To simplify this further, we can notice that the terms involving sin(k) and cos(k) appear in the numerator and the denominator with opposite signs. Thus, they cancel each other out:

= (sin(y)cos(k) + sin(y) + sin(y)cos(k) - cos(y)sin(k)) / (2cos(y)cos(k) + cos(y) + sin(y)sin(k))
= 2sin(y)cos(k) / 2cos(y)cos(k)
= sin(y) / cos(y)
= tan(y)

Therefore, we have shown that (sin x + sin y + sin z) / (cos x + cos y + cos z) = tan(y) for x, y, and z being consecutive terms of an arithmetic sequence.