Difference between revisions of "Law of Tangents"

Revision as of 01:39, 15 December 2009

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 $\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} .$

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, $\frac{\sin(s+d) - \sin(s-d)}{\sin(s+d) + \sin(s-d)} = \frac{2\cos s \sin d}{2\sin s \cos d} = \frac{\tan d}{\tan s} = \frac{\tan (A-B)/2}{\tan (A+B)/2}$ as desired. $\square$

Problems

Introductory

This problem has not been edited in. 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 12: / Line 12: @@
 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 (A-B)/2}{\tan (A+B)/2} </cmath>
-as desired.  <math>\blacksquare</math>
+as desired.  <math>\square</math>
 ==Problems==

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