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Law of tangents

In trigonometry, the law of tangents is a statement about arbitrary triangles in the plane. If two sides of a triangle are (lower-case) a and b and the angles opposite those sides are (capital) A and B, then the law of tangents states

\frac{a+b}{a-b} = \frac{\tan[\frac{1}{2}(A+B)]}{\tan[\frac{1}{2}(A-B)]}

Derivation

Start with (a + b)/(a - b). ((sin A)/a = (sin B)/b because of the law of sines):

\frac{a+b}{a-b} = \frac{a\cdot\frac{\sin A}{a} + b\cdot\frac{\sin B}{b}}{a\cdot\frac{\sin A}{a} - b\cdot\frac{\sin B}{b}}
\frac{a+b}{a-b} = \frac{\sin(A) + \sin(B)}{\sin(A) - \sin(B)} = \frac{2\sin[\frac{1}{2}(A+B)] \cdot \cos[\frac{1}{2}(A-B)]}{2\cos[\frac{1}{2}(A+B)] \cdot \sin[\frac{1}{2}(A-B)]}
(See: Trigonometric identity)
\frac{a+b}{a-b} = \frac{\sin[\frac{1}{2}(A+B)]}{\cos[\frac{1}{2}(A+B)]} \cdot \frac{\cos[\frac{1}{2}(A-B)]}{\sin[\frac{1}{2}(A-B)]}
\frac{a+b}{a-b} = \frac{\tan[\frac{1}{2}(A+B)]}{\tan[\frac{1}{2}(A-B)]}

See also

01-04-2007 01:18:14
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