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Difference between revisions of "Law of Tangents"

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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.
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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]].
  
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==Theorem==
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== Statement ==
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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>.
 
  
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==Proof==
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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
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First we can write the [[RHS]] in terms of [[sine]]s and [[cosine]]s:
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<cmath> \frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} . </cmath>
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<cmath>\frac{a-b}{a+b}=\frac{\sin (A-B)/2 \cos (A+B)/2}{\sin (A+B)/2\cos (A-B)/2}</cmath>
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We can use various sum-of-angle trigonometric identities to get:
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== Proof ==
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<cmath>\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A +\sin B}</cmath>
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By the law of sines, we have
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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]],
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<cmath>\frac{a-b}{a+b}=\frac{a/2R-b/2R}{a/2R+b/2R}</cmath>
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<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>
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where <math>R</math> is the [[circumradius]] of the triangle. Applying the law of sines again,
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<cmath>\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}</cmath>
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(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.)
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as desired.
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{{halmos}}
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By the angle addition identities,
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<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>
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as desired. <math>\square</math>
  
 
==Problems==
 
==Problems==
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{{problem}}
 
{{problem}}
 
===Intermediate===
 
===Intermediate===
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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>.
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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>
 
<div align="right">([[Mu Alpha Theta]] 1991)</div>
 
===Olympiad===
 
===Olympiad===

Latest revision as of 16:11, 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 $A$ and $B$ are angles in a triangle opposite sides $a$ and $b$ respectively, then \[\frac{a-b}{a+b}=\frac{\tan [\frac{1}{2}(A-B)]}{\tan [\frac{1}{2}(A+B)]} .\]

Proof

Let $s$ and $d$ denote $(A+B)/2$, $(A-B)/2$, respectively. By the Law of Sines, \[\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .\]

(In general, since $\frac{x}{sin X}$ 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, \[\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan [\frac{1}{2} (A-B)]}{\tan[ \frac{1}{2} (A+B)]}\] as desired. $\square$

Problems

Introductory

This problem has not been edited in. If you know this problem, please help us out by adding it.

Intermediate

In $\triangle ABC$, let $D$ be a point in $BC$ such that $AD$ bisects $\angle A$. Given that $AD=6,BD=4$, and $DC=3$, find $AB$.

Olympiad

Show that $[ABC]=r^2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$.

(AoPS Vol. 2)

See Also

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