Four trigonometric values in arithmetic progression: avoiding six orderings

slime_ranchero·Sep 30, 2026(edited)

I started checking all orders of sin, cos, tan, cot and it became a mess. Is there a symmetry that cuts the number of cases down? Problem: Does there exist an angle

\alpha \in (0, \pi / 2)

such that

\sin \alpha

\cos \alpha

\tan \alpha

and

\cot \alpha

, taken in some order, are consecutive terms of an arithmetic progression? Source: Baltic Way 2008 · MNET-0153

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3 replies

ApertureIntern·Sep 30, 2026

Replace α by π/2−α if necessary, so α≤π/4. The smallest value is sin α and the largest is cot α; only cos and tan can swap order. At α=π/4 the multiset has two distinct values repeated twice, so it cannot be four consecutive terms of an arithmetic progression.

slime_ranchero·Oct 1, 2026

For α<π/4, in either possible order, the distance cos−sin equals cot−tan: either both are one step or both are two steps.

ApertureIntern·Oct 1, 2026

Exactly. But

\cot\alpha-\tan\alpha=(\cos\alpha-\sin\alpha)(\cos\alpha+\sin\alpha)/(\sin\alpha\cos\alpha)

. Cancel the positive factor cos−sin. It would give

1=1/\sin\alpha+1/\cos\alpha>2

, a contradiction.

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