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IMO LongList 1967, Mongolia 6

Source: IMO LongList 1967, Mongolia 6

December 16, 2004
trigonometryalgebraseries summationTrigonometric IdentitiesIMO ShortlistIMO Longlist

Problem Statement

Prove the identity ∑k=0n(nk)(tan⁡x2)2k(1+2k(1−tan⁡2x2)k)=sec⁡2nx2+sec⁡nx\sum\limits_{k=0}^n\binom{n}{k}\left(\tan\frac{x}{2}\right)^{2k}\left(1+\frac{2^k}{\left(1-\tan^2\frac{x}{2}\right)^k}\right)=\sec^{2n}\frac{x}{2}+\sec^n x for any natural number nn and any angle x.x.
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