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law of tangents

(′lö əv ′tan·jəns)

(mathematics) Given a triangle with angles A, B, and C and sides a, b, c opposite these angles respectively: (a - b)/(a + b) = [tan ½(A - B)]/[tan ½(A + B)].


 
 
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Fig. 1 - A triangle.
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Fig. 1 - A triangle.
Trigonometry

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Functions
Inverse functions
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List of identities
Exact constants
Generating trigonometric tables
CORDIC

Euclidean theory

Law of sines
Law of cosines
Law of tangents
Pythagorean theorem

Calculus

The Trigonometric integral
Trigonometric substitution
Integrals of functions
Integrals of inverses

In trigonometry, the law of tangents is a statement about arbitrary triangles in the plane.

In Figure 1, a, b, and c are the lengths of the three sides of the triangle, and α, β, and γ are the angles opposite those three respective sides. The law of tangents states that

\frac{a-b}{a+b} = \frac{\tan[\frac{1}{2}(\alpha-\beta)]}{\tan[\frac{1}{2}(\alpha+\beta)]}.

The law of tangents, although not as commonly known as the law of sines or the law of cosines, is just as useful, and can be used in any case where you know either two sides and an angle, or two angles and a side.

Proof

To prove the law of tangents we can start with the law of sines:

\frac{a}{\sin{\alpha}} = \frac{b}{\sin{\beta}}.

Writing q for this common value, we get Failed to parse (unknown function\scriptstyle): \scriptstyle{a\,=\,q\sin\alpha} , Failed to parse (unknown function\scriptstyle): \scriptstyle{b\,=\,q\sin\beta} , so

\frac{a-b}{a+b} = \frac{q \sin \alpha -q\sin\beta}{q\sin\alpha+q\sin\beta} = \frac{ \sin \alpha -\sin\beta}{\sin\alpha+\sin\beta}.

Using the trigonometric identity

\sin(x) + \sin(y) = 2 \sin\left( \frac{x + y}{2} \right) \cos\left( \frac{x - y}{2} \right) \;

for Failed to parse (unknown function\scriptstyle): \scriptstyle{x\,=\,\alpha}

and Failed to parse (unknown function\scriptstyle): \scriptstyle{y\,=\,\pm\beta}
we get 
\frac{a-b}{a+b} =  \frac{   2 \sin\left( \frac{\alpha -\beta}{2} \right) \cos\left( \frac{\alpha+\beta}{2}\right)                           }{               2 \sin\left( \frac{\alpha +\beta}{2} \right) \cos\left( \frac{\alpha-\beta}{2}\right)} = {{\tan{\alpha - \beta \over 2}} \over {\tan{\alpha + \beta \over 2}}}.

See also


 
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