Difference between revisions of "Law of Tangents"

Revision as of 16:49, 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 $A$ 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, $\frac{a-b}{a+b} = \frac{\sin A - \sin B}{\sin A + \sin B} = \frac{ \sin(s+d) - \sin (s-d)}{\sin(s+d) + \sin(s-d)} .$ 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 11: / Line 11: @@
 <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>
+<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>

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