Difference between revisions of "Law of Tangents"
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| − | The '''Law of Tangents''' | + | The '''Law of Tangents''' is a useful [[trigonometric identity]] that, along with the [[law of sines]] and [[law of cosines]], can be used to determine [[angle]]s in a triangle. Note that the law of tangents usually cannot determine sides, since only angles are involved in its statement. |
| − | |||
| + | ==Theorem== | ||
| + | The law of tangents states that in triangle <math>\triangle ABC</math>, if <math>A</math> and <math>B</math> are angles of the triangle opposite sides <math>a</math> and <math>b</math> respectively, then <math>\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}</math>. | ||
| − | ==See | + | ==Proof== |
| + | First we can write the [[RHS]] in terms of [[sine]]s and [[cosine]]s: | ||
| + | <cmath>\frac{a-b}{a+b}=\frac{\sin (A-B)/2 \cos (A+B)/2}{\sin (A+B)/2\cos (A-B)/2}</cmath> | ||
| + | We can use various sum-of-angle trigonometric identities to get: | ||
| + | <cmath>\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A +\sin B}</cmath> | ||
| + | By the law of sines, we have | ||
| + | <cmath>\frac{a-b}{a+b}=\frac{a/2R-b/2R}{a/2R+b/2R}</cmath> | ||
| + | where <math>R</math> is the [[circumradius]] of the triangle. Applying the law of sines again, | ||
| + | <cmath>\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}</cmath> | ||
| + | as desired. | ||
| + | {{halmos}} | ||
| + | |||
| + | ==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]] | * [[Trigonometry]] | ||
* [[Trigonometric identities]] | * [[Trigonometric identities]] | ||
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* [[Law of Cosines]] | * [[Law of Cosines]] | ||
| − | + | [[Category:Theorems]] | |
| − | |||
[[Category:Trigonometry]] | [[Category:Trigonometry]] | ||
Revision as of 13:10, 2 February 2008
The Law of Tangents is a useful trigonometric identity that, along with the law of sines and law of cosines, can be used to determine angles in a triangle. Note that the law of tangents usually cannot determine sides, since only angles are involved in its statement.
Theorem
The law of tangents states that in triangle 
, if 
and 
are angles of the triangle opposite sides 
and 
respectively, then 
.
Proof
First we can write the RHS in terms of sines and cosines:
![\[\frac{a-b}{a+b}=\frac{\sin (A-B)/2 \cos (A+B)/2}{\sin (A+B)/2\cos (A-B)/2}\]](http://latex.artofproblemsolving.com/f/6/a/f6a354e82e824383836fc351de79bcad7cce15ee.png)
We can use various sum-of-angle trigonometric identities to get:
![\[\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A +\sin B}\]](http://latex.artofproblemsolving.com/3/6/d/36d546d755a4068f486fdf9e4d97d88238b4c155.png)
By the law of sines, we have
![\[\frac{a-b}{a+b}=\frac{a/2R-b/2R}{a/2R+b/2R}\]](http://latex.artofproblemsolving.com/3/4/1/34154a907b2bf8e505494710dc8356a632df1312.png)
where 
is the circumradius of the triangle. Applying the law of sines again,
![\[\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}\]](http://latex.artofproblemsolving.com/f/9/2/f9235368ea3caa10f3b86852de5a95ef7c506ad9.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)
.
