More Trigonometric Identities Revealed

For today there is another question from Yahoo Answers, also about trigonometric identities. This time it's not about simple ones like before, but a more complex one, like the ones I sometimes answer on super math tips. Here it is:

Prove:
(cos 4x + cos 2x) / (sin 4x + sin 2x) = cot 3x

At first glance this seems irritating and impossible, but with some creative thinking it's very possible.

Now, what do 2, 3, and 4 have in common? Of course, 3 is the average, and we're going to use that property heavily.

Since we want to have cot 3x at the end, let's change all sines and cosines to 3x + x or 3x - x and hope they cancel:
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x.

Well, they don't cancel so fast, so we'll have to expand all expressions (remember what's sin (A + B)?) and then hope that we can cancel out some terms. Let's start (I put the arguments in parentheses so you can see them better):
(cos (3x + x) + cos (3x - x)) / (sin (3x + x) + sin(3x - x)) = cot 3x
Let's start from the numerator:
cos(3x)cos(x) - sin(3x)sin(x) + cos(3x)cos(x) + sin(3x)sin(x)
2cos(3x)cos(x)

Now, the denominator:
sin(3x)cos(x) + cos(3x)sin(x) + sin(3x)cos(x) - cos(3x)sin(x)
2sin(3x)cos(x)

Now divide those two terms and see what happens:
2cos(3x)cos(x) / 2sin(3x)cos(x)
The 2 and cos(x) cancel:
cos(3x) / sin(3x) = 1/(sin(3x) / cos(3x)) = 1/tan(3x) = cot (3x)

Job complete.

Besides solving these questions, here is a challenge for you: try to solve the previous question (with the towns) using calculus. I haven't solved it myself yet (except for the geometry part, which is easy).

The first one to solve (on comments or email) will be written in the hall of fame on the blog's list (which I'll create) with a link of his choice and the right to post here any question.

Go on to solving!
Nadav

nadavs