Rational Trigonometry. Rational Trigonometry

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Milton Booth
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There are an infinite number of points on the unit circle whose coordinates are each rational numbers. Indeed every Pythagorean triple gives one!

1 There are an infinite number of points on the unit circle whose coordinates are each rational numbers. Indeed every Pythagorean triple gives one! Consider the tetrahedron with vertices (1 1 1) (1 %1 %1) (%1 1 %1) (%1 %1 1). What is the cosine of the angle formed by rays from the origin to two of these vertices? The cosine is % 1 and the sine is 3 3.

2 A rational angle is an angle whose degree measure is a rational number. Equivalently it's an angle whose radian measure is a rational number times &. It's not too hard to see that the sine and cosine of a rational angle are algebraic numbers. Toward this end one may use the angle sum formulas for sine and cosine to build up expressions for cos(3 a) cos(4 a) cos(5 a) etc. that are polynomials in cos(a). cos(x + y) = cos(x) cos(y) % sin(x) sin(y) sin(x + y) = sin(x) cos(y) + cos(x) sin(y) cos(3 a) = cos( a + a) = cos( a) cos(a) % sin( a) sin(a) = cos (a) % 1 cos(a) % ( sin(a) cos(a)) sin(a) = cos 3 (a) % cos(a) % sin (a) cos(a) = cos 3 (a) % cos(a) % 1 % cos (a) cos(a) = 4 cos 3 (a) % 3 cos(a) This process builds the Chebyshev polynomials of the first kind. cos( a) = cos (a) $ 1 cos(3 a) = 4 cos 3 (a) $ 3 cos(a) cos(4 a) = 8 cos 4 (a) $ 8 cos (a) + 1 cos(5 a) = 16 cos 5 (a) $ 0 cos 3 (a) + 5 cos(a) cos(6 a) = 3 cos 6 (a) $ 48 cos 4 (a) + 18 cos (a) $ 1 cos(7 a) = 64 cos 7 (a) $ 11 cos 5 (a) + 56 cos 3 (a) $ 7 cos(a) cos(8 a) = 18 cos 8 (a) $ 56 cos 6 (a) cos 4 (a) $ 3 cos (a) + 1 cos(9 a) = 56 cos 9 (a) $ 576 cos 7 (a) + 43 cos 5 (a) $ 10 cos 3 (a) + 9 cos(a) cos(10 a) = 51 cos 10 (a) $ 180 cos 8 (a) cos 6 (a) $ 400 cos 4 (a) + 50 cos (a) $ 1

3 Letting a = m & we see that cos(n a) = ±1 so cos(a) is a root of the appropriate n Chebyshev polynomial plus or minus 1. Example: With n = 3 we get cos(& 3) is a root of 4 x 3 % 3 x + 1 so the cosine of & 3 is algebraic. (But we knew that!) This reasoning shows cos m n & is algebraic for all rational numbers m n. Finally 1 % cos m n & is also algebraic so sin m n & is algebraic too. Which rational angles have a super-easy-to-remember sine and cosine? Specifically: Which rational angles - have sin (-) and cos (-) both rational? Here are a few such rational angles in the first quadrant: 0 & 6 & 4 & 3 &. Are there more? = 1 % = 1 = 1

4 LEMMA 1. Let b 0 be any rational number with % b 0. Define a sequence recursively by b k+1 = b k %. Then: (i) Each b k is rational. (ii) Each b k satisfies % b k. (iii) If the same value is ever attained twice say b m = b m+k then b m {% %1 0 1 }. (iv) If b m {% %1 0 1 } and m > 0 then b m%1 {% %1 0 1 }. Thus if the sequence b 0 b 1 b ever repeats a term then b 0 {% %1 0 1 }. (iii) If the same value is ever attained twice say b m = b m+k then b m {% %1 0 1 }. Proof of (iii) Let f (x) = x % and for any natural number n let f n (x) = f (f (f (4 f (x) 4 ))) be the n- fold composition of f. This is a monic polynomial of degree n with integer coefficients. Note that f k (b m ) = f k (f m (b 0 )) = f m+k (b 0 ) = b m+k = b m. This means b m is a rational root of f k (x) % x. This means that b m is an integer! (Gauss Lemma: Every rational root of a monic polynomial with integer coefficients must be an integer.)

5 LEMMA. Let - 6 and let b 0 = cos(-). Then b k = cos k - for all k. Proof: This is a simple inductive argument that makes use of the double angle formula for cosine. b k = b k%1 % b k = cos k%1 - % b k = 4 cos k%1 - % b k = cos k%1 - % 1 b k = cos k - THEOREM Suppose - is a rational angle and that cos (-) is rational. Then cos(-) and sin(-) both belong to the set 0 ±1 ± 1 ± ± 3. Proof: Write - = m n & and we suppose that cos (-) is rational (which implies sin (-) is rational). It suffices to show cos(-) belongs to the set 0 ± 1 ± ± 3 ±1. Let b 0 = cos( -). By the double angle formula b 0 = 4 cos (-) % which is rational and clearly % b 0 so this defines the sequence b 0 b 1 b of lemma 1. By lemma b k = cos m k &. The key is that the only possible remainders upon n division of an integer by n are 0 1 n % 1. So after n + 1 terms in the sequence there must be two terms that are equal. But lemma 1 then tells us that b 0 {% %1 0 1 }. This gives cos m & %1 % Or (using the double angle formula yet n again): cos m n & % 1 %1 % A few short calculations should allow you to reach the conclusion.

6 COROLLARY Suppose - is a rational angle and that cos(-) is rational. Then cos(-) belongs to the set 0 ±1 ± 1. Proof: SInce cos (-) is also rational the first theorem tells us that cos(-) 0 ± 1 ± ± 3 ±1. The only rational members of this set are 0 ±1 ± 1. Further Reading Bergen Jeffery Values of trigonometric functions Math Horizons Feb. 009 pp Jahnel Jörg When is the (co)sine of a rational angle equal to a rational number? arxiv: [math.ho]. Nivan Ivan. Irrational Numbers MAA 1956.

a. y= 5x 2 +2x 3 d. 2x+5=10 b. y= 3 x 2 c. y= 1 x 3 Learning Goal QUIZ Trigonometric Identities. OH nooooooo! Class Opener: March 17, 2015

a. y= 5x 2 +2x 3 d. 2x+5=10 b. y= 3 x 2 c. y= 1 x 3 Learning Goal QUIZ Trigonometric Identities. OH nooooooo! Class Opener: March 17, 2015 DAY 48 March 16/17, 2015 OH nooooooo! Class Opener: Find D and R: a. y= 5x 2 +2x 3 b. y= 3 x 2 c. y= 1 x 3 +2 d. 2x+5=10 Nov 14 2:45 PM Learning Goal 5.1.-5.2. QUIZ Trigonometric Identities. Mar 13 11:56