Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Derivation of Law of Sines and Cosines


Date: 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/   
    
Associated Topics:
College Triangles and Other Polygons
College Trigonometry
High School Triangles and Other Polygons
High School Trigonometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/