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Law of tangents - Wikipedia, the free encyclopedia

Law of tangents

From Wikipedia, the free encyclopedia

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

History
Usage
Functions
Inverse functions
Further reading

Reference

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 the relationship between the lengths of the three sides of a triangle and the tangents of the angles.

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 two sides and an angle, or two angles and a side are known.

[edit] Proof

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

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

We can say there's a q that equals to,

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

With this identity, we can solve for both b and a as such,

a = qsinα and b = qsinβ

Substituting in the original equation for a and b we get,

\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}

Cancelling the q's, and using the trigonometric identity

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

for \scriptstyle{x\,=\,\alpha} and \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)} = \frac{\tan[\frac{1}{2}(\alpha-\beta)]}{\tan[\frac{1}{2}(\alpha+\beta)]}

[edit] See also


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