A simple proof of the transcendence of the trigonometric functions ✩ Yuanyuan Liana, Kai Zhanga,∗ aDepartment of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi, 710129, PR China Abstract In this note, we give a simple proof that the values of the trigonometric functions at any nonzero rational number are transcendental numbers. Keywords: Irrational number, Transcendental number, Trigonometric function 2010 MSC: 11J81 The irrationality and transcendence of numbers have attracted a lot of interest, especially for the numbers related to e and π. In 1737, Euler [3] proved the irrational- ity of e. The irrationality of π was proved by Lambert [8] (see also [2, Chap. 34]) in 1761. Both proofs depend heavily on the continued fractions. In 1873, Hermite [5] gave the proof of the transcendence of e by a new technique, which involves con- structing an auxiliary function and the integral by parts. This technique has been developed extensively and all the results cited below are based on this technique. The transcendence of π was finally proved by Lindemann [9, 10] in 1882. In addition, it can be proved (see [1, 7] and [4, Theorem 48, Chap. 3]) that the exponential function ex maps rational numbers (=6 0) to irrational numbers. In 1947, Niven [11] gave a simple proof of the irrationality of π. A modified proof shows that the trigonometric functions map rational numbers (=6 0) to irrational numbers (see [12, Theorem 2.5]). With the aid of the idea of Hurwitz [6], Niven [12] gave a simple proof of the transcendence of e. It indicates that the exponential function ex maps arXiv:1912.05348v1 [math.NT] 11 Dec 2019 rational numbers (=6 0) to transcendental numbers. In this note, based on the technique of Hermite, we give a simple proof of the ✩Research is supported by NSFC 11701454. ∗Corresponding author. Email addresses: [email protected] (Yuanyuan Lian), [email protected] (Kai Zhang) Preprint submitted to December 12, 2019 transcendence of trigonometric functions. That is, the trigonometric functions map rational numbers (=6 0) to transcendental numbers. Theorem 0.1. For any rational number r =06 , cos r is a transcendental number. Proof. Since cos r = cos(−r), we can assume that r is positive and write r = s/t where s, t are relatively prime positive integers. We use a proof by contradiction. Suppose that cos r is not a transcendental number. Then there exist a positive integer m and rational numbersa ¯j (0 ≤ j ≤ m) such that m j a¯j cos r =0. Xj=0 From the product-to-sum formulas for trigonometric functions, there exist rational numbers aj (0 ≤ j ≤ m) such that m aj cos jr =0. (0.1) Xj=0 By multiplying an integer, we may assume that aj (0 ≤ j ≤ m) are all integers. If a0 =6 0, let t4mp+2p−2x2p−2(x2 − r2)2p(x2 − 4r2)2p ··· (x2 − m2r2)2p f(x)= , (0.2) (2p − 2)! where p is a positive prime to be specified later. If a0 = 0 and aj0 =6 0 for some 1 ≤ j0 ≤ m, we take f to be 4mp+2p−2 2p−2 t f (x)f (x) ··· f − (x)(x − j r) f (x) ··· f (x) f(x)= 0 1 j0 1 0 j0+1 m (2p − 2)! 2 2 2p where fk(x) = ((x − j0r) − (kr − j0r) ) for k =6 j0. In both cases, the following arguments are almost the same. Thus, we only give the proof for the case a0 =6 0. For 0 <x<mr, we have t4mp+2p−2(mr)4mp+2p−2 |f(x)| < . (0.3) (2p − 2)! It can also be easily verified that for any k ≥ 0, f k(0) is an integer and f (2k+1)(0) = 0 (see [12, Chapter 2.2]). 2 By the elementary calculus, for any positive integer j, jr ∞ ∞ ∞ f(x) sin(jr−x)dx = (−1)kf (2k)(jr)+ (−1)kf (2k)(0) cos jr+ (−1)kf (2k+1)(0) sin jr. Z 0 Xk=0 Xk=0 Xk=0 Note that f (2k+1)(0) = 0 for any k. Hence, jr ∞ ∞ f(x) sin(jr − x)dx = (−1)kf (2k)(jr)+ (−1)kf (2k)(0) cos jr. Z 0 Xk=0 Xk=0 Then, by(0.1), m jr m ∞ a f(x) sin(jr − x)dx = (−1)ka f (2k)(jr). (0.4) j Z j Xj=0 0 Xj=0 Xk=0 The right hand in above equation is an integer. Moreover, by(0.2), f (2k)(jr) is divisible by p for all k and j with one exception: f (2p−2)(0) = s4mpt2p−2(m!)4p, if we choose p > max {m, s, t}. Next, by taking p > |a0|, the right hand of(0.4) (2p−2) consists of a sum of multiples of p with one exception, namely a0f (0). Hence, the right hand of(0.4) is a non-zero integer. However, by(0.3), the left hand of(0.4) satisfies m m jr t4mp+2p−2(m2r2)2mp+2p−2 aj f(x) sin(jr − x)dx ≤ |aj|· jr · < 1, Z0 (2p − 2)! Xj=0 Xj=0 provided p is chosen sufficiently large. Thus we have a contradiction, and the theo- rem is proved. Corollary 0.2. The trigonometric functions are transcendental at non-zero rational values of the arguments. Proof. Because of cos 2r =1 − sin2 r and cos2r = (1 − tan2 r)/(1+tan2 r), sin r and tan r are transcendental numbers for any rational number r =6 0. 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