Difference between revisions of "Law of Tangents"
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| − | <math>\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a | + | 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 [[angle]]s or sides in a [[triangle]]. |
| + | |||
| + | == Statement == | ||
| + | |||
| + | 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 [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . </cmath> | ||
| + | |||
| + | == Proof == | ||
| + | |||
| + | 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> | ||
| + | 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 [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]} </cmath> | ||
| + | as desired. <math>\square</math> | ||
| + | |||
| + | ==Problems== | ||
| + | ===Introductory=== | ||
| + | {{problem}} | ||
| + | ===Intermediate=== | ||
| + | In <math>\triangle ABC</math>, let <math>D</math> be a point in <math>BC</math> such that <math>AD</math> bisects <math>\angle A</math>. Given that <math>AD=6,BD=4</math>, and <math>DC=3</math>, find <math>AB</math>. | ||
| + | <div align="right">([[Mu Alpha Theta]] 1991)</div> | ||
| + | ===Olympiad=== | ||
| + | Show that <math>[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}</math>. | ||
| + | |||
| + | <div align="right">(AoPS Vol. 2)</div> | ||
| + | ==See Also== | ||
| + | * [[Trigonometry]] | ||
| + | * [[Trigonometric identities]] | ||
| + | * [[Law of Sines]] | ||
| + | * [[Law of Cosines]] | ||
| + | |||
| + | [[Category:Theorems]] | ||
| + | [[Category:Trigonometry]] | ||
Revision as of 17:49, 2 March 2017
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,
![\[\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]}\]](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 
.
(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)
