Proving limit of trigonometric functions

[exraven]

Homework Statement

Prove:
lim (n→0) {(sin n cos n) / (ⁿ√(1 - tan² 2n + tan⁴ 2n - tan⁶ 2n + ...))} = 1

Homework Equations

The Attempt at a Solution

The denominator is an infinite geometric series, using the sum formula of an infinite geometric series, I simplify the limit:
lim (n→0) {(sin n cos n) / (ⁿ√(1 - tan² 2n + tan⁴ 2n - tan⁶ 2n + ...))}
= lim (n→0) {(sin n cos n) / (ⁿ√(1 / (1 + tan² 2n)))}

After a few tries, i can't figure out what to do with the ⁿ√, rationalizing the denominator won't work because the ⁿ√ will stuck in the numerator. L'hopital is not allowed because this is a high school limit problem. Could anyone give me some hints to solve this ? Any help would be much appreciated.

 

[lanedance]

have a look at the first formula here
http://en.wikipedia.org/wiki/List_o...Double-.2C_triple-.2C_and_half-angle_formulae

also have a look at some of these
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Related_identities

 

[exraven]

sorry for the late reply, i was struggling to solve the problem. It seems that trigonometric identities won't work, the closest one is using the euler's formula, the result is 0. It fail to prove the problem though, maybe it is a false problem after all.

 

[Char. Limit]

Part of the problem is that the limit above doesn't equal 1. And the best way I can think of to do this is to find just what the infinite series converges to. Although I'm not sure if that's in the scope of the problem.