Difference between revisions of "Law of Tangents"
(→Proof) |
|||
| Line 4: | Line 4: | ||
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 == | ||
Revision as of 22:23, 25 October 2013
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)
By the angle addition identities,
![\[\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2}\]](http://latex.artofproblemsolving.com/d/7/3/d73ff95d36c6ccf90f648760ef6272033ce68061.png)
as desired. 
Problems
Introductory
This problem has not been edited in. Help us out by adding it.
Intermediate
In 
, let 
be a point in 
such that 
bisects 
. Given that 
, and 
, find 
.
(Mu Alpha Theta 1991)
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)
.
(AoPS Vol. 2)
