Derivation of Law of Sines and CosinesDate: 11/02/97 at 14:58:12 From: Yuan Tao Subject: Derivation of law of Sines and Cosines How do you derive the law of sines and the law of cosines? Date: 11/03/97 at 14:31:23 From: Doctor Pete Subject: Re: Derivation of law of Sines and Cosines Generally, there are several ways to prove the Law of Sines and the Law of Cosines, but I will provide one of each here: Let ABC be a triangle with angles A, B, C and sides a, b, c, such that angle A subtends side a, etc. Theorem (Law of Sines). Sin[A]/a = Sin[B]/b = Sin[C]/c. Proof. Draw the perpendicular (altitude) from B to side b, and suppose this has length h. Then note that Sin[C] = h/a, or h = a Sin[C]. Similarly, Sin[A] = h/c, or h = c Sin[A]. Therefore, a Sin[C] = c Sin[A] or Sin[A]/a = Sin[C]/c, since neither a nor c has length 0. Clearly, if we were to draw another altitude h', say, from A to a, we would find by a similar argument that Sin[B]/B = Sin[C]/c, and hence the result follows. Theorem (Law of Cosines). c^2 = a^2 + b^2 - 2ab Cos[C]. Proof. Place triangle ABC on a Cartesian coordinate system such that angle C is at the origin and length a lies on the x-axis. Then length b is the other ray from the origin. We can easily identify the coordinates of two of the vertices: Vertex C lies at (0,0), and vertex B lies at (a,0). We must identify vertex A, which lies somewhere in quadrant I or II (since angle C < 180 degrees). But the x-coordinate of A is b Cos[C], as can be seen by drawing the perpendicular from A to the x-axis. Similarly, the y-coordinate is b Sin[C]. Hence A lies at (b Cos[C], b Sin[C]). Therefore, the length of c is given by the distance between A and B, which is by the distance formula c = Sqrt[(a - b Cos[C])^2 + (b Sin[C])^2] or upon squaring both sides, c^2 = (a - b Cos[C])^2 + (b Sin[C])^2 = a^2 - 2ab Cos[C] + b^2 Cos[C]^2 + b^2 Sin[C]^2 = a^2 + b^2 (Cos[C]^2 + Sin[C]^2) - 2ab Cos[C]. But Cos[C]^2 + Sin[C]^2 = 1 for any angle C, and therefore c^2 = a^2 + b^2 - 2ab Cos[C]. I hope that was understandable. To really get a feel for the proofs, draw out the diagrams. Clearly, the Law of Cosines is a more difficult proof, but it is really a generalization of the Pythagorean Theorem, since in the right-angle case (C = 90), Cos[C] = 0 and we obtain c^2 = a^2 + b^2. This is why we often see the Law of Cosines written as c^2 = ... instead of the equally valid form a^2 = b^2 + c^2 - 2bc Cos[A]. Another little tidbit of interest is that the proof of the Law of Sines can be taken in a slightly different direction to show that the area of a triangle is (ab Sin[C])/2 = (bc Sin[A])/2 = (ac Sin[B])/2. There's also a Law of Tangents, but I don't quite remember the formula. It's not commonly used, and it really amounts to just a bunch of trigonometric manipulation, so it's not that exciting. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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