Difference between revisions of "Law of Tangents"

Revision as of 23:04, 2 February 2008

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. $\blacksquare$

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)

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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.
+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]].
-==Theorem==
+== Statement ==
-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>.
-==Proof==
+If <math>A</math> and <math>B</math> are angles in a triangle opposite sides <math>a</math> and <math>b</math> respectively, then
-First we can write the [[RHS]] in terms of [[sine]]s and [[cosine]]s:
+<cmath> \frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2} . </cmath>
-<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:
+== Proof ==
-<cmath>\frac{a-b}{a+b}=\frac{\sin A-\sin B}{\sin A +\sin B}</cmath>
-By the law of sines, we have
+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{a/2R-b/2R}{a/2R+b/2R}</cmath>
+<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>
-where <math>R</math> is the [[circumradius]] of the triangle. Applying the law of sines again,
+By the angle addition identities,
-<cmath>\frac{a-b}{a+b}=\frac{\tan (A-B)/2}{\tan (A+B)/2}</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 (A-B)/2}{\tan (A+B)/2} </cmath>
-as desired.
+as desired.  <math>\blacksquare</math>
-{{halmos}}
 ==Problems==

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