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Hi, Scythian, I really love this question. The identity is utterly impressive. The key is to repeatedly use the double angle formula, 1 + cos(2 x) = 2 cos(x)², this gives

2 cos(x) = [2 cos(x/2) ]² − 2, ...................(1)

2 cos(x/2) = [2 cos(x/4) ]² − 2, ...................(2)

2 cos(x/4) = [2 cos(x/8) ]² − 2,....................(3)

....

2 cos(x/2^{n−1}) = [2 cos(x/2^n) ]² − 2,.......(n)

Plug (n) into (n−1), ...., (3) into (2), (2) into (1), and for the final one (n), we use cos(x) = 1 − x²/2 (approximately)

[2 cos(x/2^n)]² = [ 2 (1 − x²/2^{2n}/2) ]² = [ x²/2^{2n} − 2]²,

we have the desired result. (To make it exact, we need to replace x²/2^{2n} by [2 sin(x/2^{n+1})]² ).

This is a beautiful identity, and it is quite practical as you mentioned. May I ask did you discover it yourself? If so, I guess you may be a genius like Euler :-) who proved the famous infinite product

sin(x)/x = cos(x/2) cos(x/4) cos(x/8) ...

where the other double angle formula, sin(2 x) = 2 sin(x) cos(x) is used.

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Also, we can derive a hyperbolic version

(((((( (1/2^(2n))x² + 2)² − 2)² − 2)² − 2)² − 2)² − 2)² − 2... = 2 cosh(x). The proof is similar.

Another interesting point, the recursion is actually a logistic map x_{n+1} = (x_n)² − R, with a value R=2 which makes the map chaotic (it corresponds to x_{n+1} = 4 x_n(1 − x_n) in terms of the original logistic map, see http://en.wikipedia.org/wiki/Logistic_map ). This means even if we start with two very close numbers as the initial values (the closeness is measured in terms of |x− x'|/2^n), the final results can be very different. This is quite expected, because the cosine function oscillates between −1 and 1 wildly at a large x. There are many other things we can say about this identity.

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If we reversely use the double angle formula, i.e.,

2 cos(x/2) = √(2 + 2 cos(x)) and use π/2 for the last one, then

2 cos(π/2^{n+1}) = √( 2+ √( 2 + √(2 + ... √2)))

where n is the number of √.

Combine this with Euler's infinite product (mentioned above) we have an expansion of 2/π (Viète's formula).

2/π = √2/2 {[√( 2+ √2)]/2} {[√( 2+ √(2+√2))]/2}

see http://en.wikipedia.org/wiki/Vi%C3%A8te's_formula

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Now let me derive something of the sort for the tangent. For a large N, if t = θ/N, we have the following continued fraction,

tan θ ≈ 1/(−t + 1/(t + 1/(−t + 1/(t + ...1/(−t + 1/t )))))

where the number of minus signs is N.

The accuracy of the above version is only O(1/N), the following version converges better:

tan θ ≈ 1/(−t + s/(t + 1/(−t + s/(t + ... + 1/(−t + s/t )))))

where s = 1 + t². For example, for N=4, θ=π/4, I calculate tan(π/4) ≈ 0.98, which is fairly close to 1.0.

This is derived from the following observation

tan(x + δ) = (tan x + tan δ) / (1 − tan x tan δ)

= 1/(− tan δ + sec²δ/(tan δ + tan x))

Now if we call t = tan δ ≈ δ, s = sec²δ ≈ 1 + δ², and iterate the process N times, the desired result is reached.

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To Scythian: even the formula exists in literature, it is still quite amazing that you discover it independently. :-)

It is quite difficult (in fact, impossible) to prove that something is false. However, if you would like a similar "false" proof that 2 = 1, here it goes: Let x = 1 and y = 1. Clearly, x = y, right? let's multiply both sides by x from this point. This gives us x^2 = xy. Now, subtract y^2 both sides, to get x^2 - y^2 = xy - y^2. Factor both sides. (x - y)(x + y) = y(x - y). Look at how both sides have (x - y). That means we can effectively cancel them out by dividing both sides by (x - y). This leaves us with x + y = y Remember that x = 1 and y = 1, so if we back-substitute, we get 1 + 1 = 1 or 2 = 1 I'm sure the "proof" for 2 + 2 = 5 would be similar. Of course, being a false proof, there must be something wrong with the way it was conducted. Can you figure out exactly what that is?

Do I have to? Couldn't I just say I proved it and let it go at that? The last polynomial function that I tried to build came in a kit. It took me over a week to put it together.