Proving Laws of Sines, CosinesDate: 10/01/2003 at 04:19:53 From: Michael Subject: Trigonometry Hi, I am having a little trouble following the logic of the proofs of the sine and cosine laws, a/SinA = b/SinB = c/SinC and a^2 = b^2 + c^2 - 2bcosA Can you help? Date: 10/09/2003 at 07:28:29 From: Doctor Jerry Subject: Re: Trigonometry Hi Michael, There are several ways of proving these laws. I'll give you the proofs that I remember. For the law of cosines, I'll use a coordinate-system approach. A general triangle can be plotted as follows: Put vertex C at the origin, put vertex B on the positive x-axis at (a,0), and put vertex A anywhere above the x-axis. I prefer to put it in the second quadrant, above and to the left of C. The angle C is in "standard position" and so the point A will have coordinates (b*cos(C),b*sin(C)). Now calculate the square of the distance from A to B. Because it is opposite angle C we call it c^2; from the distance formula we have c^2 = (b*cos(C)-a)^2 + (b*sin(C)-0)^2 Expanding, c^2 = b^2*cos^2(C) - 2*a*b*cos(C) + a^2 + b^2*sin^2(C). The terms with b^2 can be combined: c^2 = b^2(cos^2(C) + sin^2(C)) - 2*a*b*cos(C) + a^2 c^2 = a^2 + b^2 - 2*a*b*cos(C). This is the law of cosines. For the law of sines, draw a triangle with all angles less than 90 deg. Other triangles can be handled using the same ideas, but it's a little harder to describe in words. Label the vertices A, B, and C. Label the sides opposite these vertices a, b, and c. Choose a vertex, say A. Draw an altitude from A to the opposite side. Let D be the point of intersection. Let h be the length of AD. Then sin(C) = h/b and sin(B) = h/c. Solve these two equations for h and set the results equal: b*sin(C) = c*sin(B). By symmetry, we can also show that a*sin(B) = b*sin(A) and c*sin(A) = a*sin(C). From these three equations we can see that sin(A)/a = sin(B)/b = sin(C)/c. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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