In this note we exploit recurrences of integrals to give new elementary proofs of the irrationality of tan r for r ∈ Q \ {0} and cos r for r 2 ∈ Q \ {0}. We also discuss applications of our technique to simpler irrationality proofs such as those for π, π 2 , and certain values of exponential and hyperbolic functions.1 Irrationality of tan r for r ∈ Q \ {0}.For a nonzero rational r, the irrationality of tan r was first proved by J. H. Lambert in 1761 by means of continued fractions [1, pp. 129-146]. We now present a new direct proof using a recurrence for an integral.
Theorem 1. tan r is irrational for nonzero rational r.Proof. The irrationality of π will be a by-product of this proof, so we start by supposing that r ∈ Q \ {kπ : k ∈ Z}. Write r = a/b with a, b ∈ Z and assume that tan(r/2) = p/q with p, q ∈ Z. For n ≥ 0, let f n (x) = (rx − x 2 ) n /n! and I n = r 0 f n (x) sin x dx. Then b n I n → 0 as n → ∞, I 0 = 1 − cos r, and I 1 = 2(1 − cos r) − r sin r. Integrating by parts twice, we get that for n ≥ 2,By inducting on n using (1), we see that for n ≥ 0, I n = u n (1 − cos r) + v n sin r, where u n and v n are polynomials in r with integer coefficients and degrees at most n. Moreover, if two consecutive terms of the sequence I n are 0, then (1) forces all terms of I n to be 0, and in particular I 0 = 0, a contradiction. Hence I n has infinitely many nonzero terms. Therefore, we can pick a large enough n so that b n q csc rI n = b n q[u n tan(r/2) + v n ] is a nonzero integer in (−1, 1), a contradiction.Notice that tan(π/4) = 1, so π/2 / ∈ Q, which implies that Q \ {kπ : k ∈ Z} = Q \ {0}. Thus we have proved that tan(r/2) / ∈ Q for all r ∈ Q \ {0}.A closer inspection of our proof reveals that u n and v n /r are polynomials in r 2 . So a slightly stronger conclusion can be squeezed out of the proof, namely that (tan r)/r is irrational whenever r 2 ∈ Q \ {0}. This stronger result was first established through a different elementary approach by Inkeri [2].