Spherical trigonometry

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1 Spherical trigonometry 1 The spherical Pythagorean theorem Proposition 1.1 On a sphere of radius, any right triangle AC with C being the right angle satisfies cos(c/) = cos(a/) cos(b/). (1) Proof: Let O be the center of the sphere, we may assume its coordinates are (0, 0, 0). We may rotate the sphere so that A has coordinates OA = (, 0, 0) and C lies in the xy-plane, see Figure 1. otating z y O(0, 0, 0) C A(, 0, 0) x Figure 1: The Pythagorean theorem for a spherical right triangle around the z axis by β := AOC takes A into C. The edge OA moves in the xy-plane, by β, thus the coordinates of C are OC = ( cos(β), sin(β), 0). Since we have a right angle at C, the plane of OC is perpendicular to the plane of OAC and it contains the z axis. An orthonormal basis of the plane of OC is given by 1/ OC = (cos(β), sin(β), 0) and the vector OZ := (0, 0, 1). A rotation around O in this plane by α := OC takes C into : O = cos(α) OC + sin(α) OZ = ( cos(β) cos(α), sin(β) cos(α), sin(α)). Introducing γ := AO, we have OA O cos(γ) = 2 = 2 cos(α) cos(β) 2. The statement now follows from α = a/, β = b/ and γ = c/. To prove the rest of the formulas of spherical trigonometry, we need to show the following. 1

2 Proposition 1.2 Any spherical right triangle AC with C being the right angle satisfies = sin ( ) a sin ( ) c and (2) cos(a) = tan ( ) b tan ( c ). (3) Proof: After replacing a/, b/ and c/ with a, b, and c we may assume = 1. This time we rotate the triangle in such a way that OC = (0, 0, 1), A is in the xz plane and is in the yz-plane, see Figure 2. A rotation around O in the xz plane by b = AOC takes C into A, thus we have z C(0, 0, 1) A O(0, 0, 0) y x Figure 2: Computing the sines and cosines in a spherical right triangle OA = (sin(b), 0, cos(b)). Similarly, a rotation around O in the yz-plane by a = OC takes C into, thus we have O = (0, sin(a), cos(a)). The angle A is between OA O and OA OC. Here OA O = ( cos(b) sin(a), sin(b) cos(a), sin(a) sin(b)) and OA OC = (0, sin(b), 0). The length of OA O is OA O sin(c) = sin(c), the length of OA OC is sin(b). To prove (2) we use the fact that ( OA O) ( OA OC) = OA O OA OC. (4) 2

3 Since ( cos(b) sin(a), sin(b) cos(a), sin(a) sin(b)) (0, sin(b), 0) = (sin(a) sin 2 (b), 0, sin(b) cos(b) sin(a)) the left hand side of (4) is sin 2 (a) sin 4 (b) + sin 2 (b) cos 2 (b) sin 2 (a) = sin(b) sin(a) sin 2 (b) + cos 2 (b) = sin(b) sin(a). The right hand side of (4) is is sin(b) sin(c). Thus we have yielding (2). To prove (3) we use the fact that sin(b) sin(a) = sin(b) sin(c), ( OA O) ( OA OC) = OA O OA OC cos(a). (5) The left hand side is sin 2 (b) cos(a), the right hand side is sin(b) sin(c) cos(a). Thus we obtain sin 2 (b) cos(a) = sin(b) sin(c) cos(a), yielding cos(a) = sin(b) cos(a) sin(c) Equation (3) now follows from Proposition 1.1. = tan(b) tan(c) cos(a) cos(b). cos(c) 2 General spherical triangles To prove the spherical laws of sines and cosines, we will use the Figure 3. c h a A b 1 1 b 2 C Figure 3: A general spherical triangle Theorem 2.1 (Spherical law of sines) Any spherical triangle satisfies sin(a/) = sin() sin(b/) = sin(c) sin(c/). 3

4 Proof: Applying (2) to the right triangle A 1 yields = sin(h/) sin(c/). This equation allows us to express sin(h/) as follows: sin(h/) = sin(c/). Similarly, applying (2) to the right triangle C 1 allows us to write Therefore we have sin(h/) = sin(c) sin(a/). sin(a/) sin(c/) = sin(c) sin(a/), since both sides equal sin(h/). Dividing both sides by sin(a/) sin(c/) yields sin(a/) = sin(c) sin(c/). The equality sin(a/) = may be shown in a completely similar fashion. sin() sin(b/) Theorem 2.2 (Spherical law of cosines) Any spherical triangle satisfies cos(a/) = cos(b/) cos(c/) + sin(b/) sin(c/) cos(a). Proof: Applying (1) to the right triangle 1 C yields cos(a/) = cos(b 2 /) cos(h/) Let us replace b 2 with b b 1 in the above equation. After applying the formula cos(x y) = cos(x) cos(y) + sin(x) sin(y) we obtain cos(a/) = cos(b/) cos(b 1 /) cos(h/) + sin(b/) sin(b 1 /) cos(h/). Applying (1) to the right triangle 1 A we may replace both occurrences of cos(h/) above with cos(c/)/ cos(b 1 /) and obtain cos(a/) = cos(b/) cos(c/) + sin(b/) sin(b 1 /) cos(c/) cos(b 1 /), that is, cos(a/) = cos(b/) cos(c/) + sin(b/) sin(c/) tan(b 1/) tan(c/). Finally, (3) applied to the right triangle 1 A allows replacing tan(b 1 /)/ tan(c/) with cos(a). 4

5 To obtain the spherical law of cosines for angles, we may apply the preceding theorem to the polar triangle of the triangle AC. This one has sides a = (π A), b = (π ) and c = (π C) and angles A = π a/, = π b/ and C = π c/. The spherical law of cosines for the triangle A C states cos(a /) = cos(b /) cos(c /) + sin(b /) sin(c /) cos(a ), that is, cos(π A) = cos(π ) cos(π C) + sin(π ) sin(π C) cos(π a/). Using cos(π x) = cos(x) and sin(π x) = sin(x), after multiplying both sides by ( 1) we obtain cos(a) = cos() cos(c) + sin() sin(c) cos(a/). (6) eferences [1] D. oyster, Non-Euclidean Geometry and a Little on How We Got There, Lecture notes, May 7,

Trig Identities, Solving Trig Equations Answer Section

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