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A geometrical proof of sum of cos nϕ ∗ La´szló Németh 7 1 0 Abstract 2 n In this article, we present a geometrical proof of sum of cos(cid:96)ϕ where (cid:96) goes from 1 a up to m. Although there exist some summation forms and the proofs are simple, they J use complex numbers. Our proof comes from a geometrical construction. Moreover, 4 2 from this geometrical construction we obtain an other summation form. MSC: 11L03, Key Words: Lagrange’s trigonometric identities, sum of cosnϕ. ] O H 1 Introduction . h t a The Lagrange’s trigonometric identities are well-known formulas. The one for sum of cos(cid:96)ϕ m (ϕ ∈ (0,2π)) is [ 1 (cid:88)m 1 (cid:18)sin(m+ 1)ϕ (cid:19) v cos(cid:96)ϕ = 2 −1 . (1) 6 2 sin 1ϕ 6 (cid:96)=1 2 0 7 The proof of equation (1) is based on the theorem of the complex numbers in all the books, 0 articles and lessons at the universities ([1], [2]). In the following we give a geometrical . 1 constructionwhichimpliestheformula(1)andusingcertaingeometricalpropertiesweobtain 0 an other summation formula without half angles (ϕ ∈ (0,2π),ϕ (cid:54)= π) 7 1 : m (cid:18) (cid:19) v (cid:88) 1 sin(m+1)ϕ+sinmϕ cos(cid:96)ϕ = −1 . (2) i X 2 sinϕ (cid:96)=1 r a 2 Geometrical construction Let x and e be two lines with the intersection point A . Let the angle of them is α as x is 0 rotated to e (Figure 1). Let the point A be given on x such that the distance between the 1 points A and A is 1. Let the point A be on the line e such that the distance of A and 0 1 2 1 A is also equal to 1 and A (cid:54)= A if α (cid:54)= π/2 and α (cid:54)= 3π/2. Then let the new point A 2 2 0 3 be on the line x again such that A A = 1 and A (cid:54)= A if it is possible. Recursively, we 2 3 3 1 can define the point A ((cid:96) ≥ 2) on one of the lines x or e if (cid:96) is odd or even, respectively, (cid:96) where A A = 1 and A (cid:54)= A if it is possible. Figure 1 shows the first six points and (cid:96)−1 (cid:96) (cid:96) (cid:96)−2 ∗The final publication is available at Ne´meth, L., A geometrical proof of sum of cosnϕ, Studies of the University of Zˇilina, Mathematical Series, 27 (2015) 63-66. 1 2 Figure 2 shows some general points. We can easily check that the rotation angels at vertices A ((cid:96) ≥ 1) between the line e (or the axis x) and the segments A A or A A are ((cid:96)−1)α (cid:96) (cid:96)−1 (cid:96) (cid:96) (cid:96)+1 or ((cid:96) + 1)α, respectively, as the triangles A A A are isosceles. (The angle iα can be (cid:96)−1 (cid:96) (cid:96)+1 larger the π/2, even larger than 2π. The vertices A can be closer to A then A – see (cid:96) 0 (cid:96)−2 Figure 3.) If A is on the line e we obtain a similar geometric construction. In that case 1 those points A are on line x which have even indexes. (cid:96) Figure 1: First seven points of the geometrical construction Let A be the origin and the line x is the axis x. Then the equation of the line e is 0 cosα·y = sinα·x. Let A(cid:48) be the orthogonal projection of A ∈ e (n ≥ 2) onto the axis x n n then from right angle triangle A A(cid:48) A the coordinates of the points A (see Figure 1) are 0 n n n x (α) = cotαsinnα, n (3) y (α) = sinnα. n If α = π/2 and α = 3π/2 then all the points A coincide the points A or A , so in the n 0 1 following we exclude this cases. The parametric equation system of the orbits of the points A ∈ e can be given by the n help of the Chebyshev polynomial too (for more details and for some figures of orbits, see in [3]). The equation system is x (α) = cosα U (cosα), n n−1 (4) y (α) = sinα U (cosα), n n−1 where α goes from 0 to 2π and U (x) is a Chebyshev polynomial of the second kind [4]. n−1 Let A ∈ x and n be even so that n = 2k+2. We take the orthogonal projections of the 1 segments A A ((cid:96) = 1,2,...,n) onto the line x (see Figure 1 and 2). Then we realize (cid:96)−1 (cid:96) x (α) = 1+2(cos2α+cos4α+···+cos2kα)+cos(2k +2)α, (5) n on the other hand, from (3) or from (4) we have sin(2k +2)α x (α) = cosα . (6) n sinα Comparing (5) and (6) we obtain k (cid:88) sin(2k +2)α 1+2 cos2(cid:96)α+cos(2k +2)α = cosα . (7) sinα (cid:96)=1 3 Figure 2: General points of the geometrical construction Figure 3: General points Using the addition formula for cosine we receive from (7) that k (cid:18) (cid:19) (cid:88) 1 sin(2k +2)α cos2(cid:96)α = cosα −cos(2k +2)α−1 (8a) 2 sinα (cid:96)=1 (cid:18) (cid:19) 1 cosαsin(2k +2)α−sinαcos(2k +2)α = −1 (8b) 2 sinα (cid:18) (cid:19) 1 sin((2k +2)−1)α = −1 (8c) 2 sinα (cid:18) (cid:19) 1 sin(2k +1)α = −1 . (8d) 2 sinα If ϕ = 2α then (cid:88)k 1 (cid:18)sin(k + 1)ϕ (cid:19) cos(cid:96)ϕ = 2 −1 . (9) 2 sin 1ϕ (cid:96)=1 2 3 Other summation form In this section, we give an other summation form for the cosines without half angles by the help of the defined geometrical construction. Now let us take the orthogonal projection of the segments A A ((cid:96) = 1,2,...,n) onto (cid:96)−1 (cid:96) the line e (see Figure 1 and 2) and summarize them for all (cid:96) from 1 to 2k +1. (The sum is equal to x(α) if n = 2k +1 and A ∈ e.) Now we gain a similar equation to (7), namely 1 4 sin(2k +1)α 2(cosα+cos3α+···+cos(2k −1)α)+cos(2k +1)α = cosα . sinα With analogous calculation to (8) we obtain k (cid:18) (cid:19) (cid:88) 1 sin(2k +1)α cos(2(cid:96)−1)α = cosα −cos(2k +1)α (10a) 2 sinα (cid:96)=1 (cid:18) (cid:19) 1 sin2kα = . (10b) 2 sinα Summing equations (8d) and (10b), we have k k (cid:18) (cid:19) (cid:88) (cid:88) 1 sin(2k +1)α sin2kα cos2(cid:96)α+ cos(2(cid:96)−1)α = + −1 , 2 sinα sinα (cid:96)=1 (cid:96)=1 and finally if m = 2k and ϕ = α we obtain formula (2). References [1] Mun˜iz, E.O., A Method for Deriving Various Formulas in Electrostatics and Electro- magnetism Using Lagrange’s Trigonometric Identities. American Journal of Physics 21 (2): 140 (February 1953). [2] Jeffrey, A. – Dai, H-h., (2008). ”Section 2.4.1.6” (p.129.) Handbook of Mathematical Formulas and Integrals (4th ed.). Academic Press. ISBN 978-0-12-374288-9. [3] Németh, L., A new type of lemniscate, NymE SEK Tudomańyos Ko¨zlemények XX. Természettudomańyok 15. Szombathely, (2014), 9-16. [4] Rivlin, T.J., Chebyshev polynomials, New York Wiley (1990).