Difference between revisions of "Law of Tangents"
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If <math>A</math> and <math>B</math> are angles in a triangle opposite sides <math>a</math> and <math>b</math> respectively, then | If <math>A</math> and <math>B</math> are angles in a triangle opposite sides <math>a</math> and <math>b</math> respectively, then | ||
| − | <cmath> \frac{a-b}{a+b}=\frac{\tan (A-B) | + | <cmath> \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . </cmath> |
== Proof == | == Proof == | ||
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Let <math>s</math> and <math>d</math> denote <math>(A+B)/2</math>, <math>(A-B)/2</math>, respectively. By the [[Law of Sines]], | Let <math>s</math> and <math>d</math> denote <math>(A+B)/2</math>, <math>(A-B)/2</math>, respectively. By the [[Law of Sines]], | ||
<cmath> \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . </cmath> | <cmath> \frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} . </cmath> | ||
| + | |||
| + | (In general, since <math>\frac{x}{sin X}</math> is constant in a triangle, any ratio of linear combinations applied to lengths of sides is equal to the ratio of the same linear combinations applied to the sines of the angles of the same sides.) | ||
| + | |||
By the angle addition identities, | By the angle addition identities, | ||
| − | <cmath> \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B) | + | <cmath> \frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]} </cmath> |
as desired. <math>\square</math> | as desired. <math>\square</math> | ||
Latest revision as of 15:11, 21 February 2023
The Law of Tangents is a rather obscure trigonometric identity that is sometimes used in place of its better-known counterparts, the law of sines and law of cosines, to calculate angles or sides in a triangle.
Statement
If 
and 
are angles in a triangle opposite sides 
and 
respectively, then
![\[\frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} .\]](http://latex.artofproblemsolving.com/2/b/1/2b1865930dc8faf1d675296fcdb71180500ec650.png)
Proof
Let 
and 
denote 
, 
, respectively. By the Law of Sines,
![\[\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .\]](http://latex.artofproblemsolving.com/5/5/6/55614e64015fcdcc27a9a7a5b469e537b6f3bf16.png)
(In general, since 
is constant in a triangle, any ratio of linear combinations applied to lengths of sides is equal to the ratio of the same linear combinations applied to the sines of the angles of the same sides.)
By the angle addition identities,
![\[\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]}\]](http://latex.artofproblemsolving.com/2/4/f/24f890330a8cd6741bac8d505a4e0e3e409a827c.png)
as desired. 
Problems
Introductory
This problem has not been edited in. If you know this problem, please help us out by adding it.
Intermediate
In 
, let 
be a point in 
such that 
bisects 
. Given that 
, and 
, find 
.
Olympiad
Show that ![$[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$](http://latex.artofproblemsolving.com/8/3/2/832e3595d44124851e7eefe05b5b14b46ba8980b.png)
.
