Difference between revisions of "Law of Tangents"

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 $AB$ .

(Mu Alpha Theta 1991)

Olympiad

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

(AoPS Vol. 2)

@@ Line 1: / Line 1: @@
-<math>\frac{a-b}{a+b}=\frac{\tan(a-b)}{\tan(a+b)}</math>
+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>
+(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,
+<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]]

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