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Technically, the existence of the tangent half-angle formulae stems from the fact that the Circle is an Algebraic Curve of Genus 0. One then expects that the 'circular functions' should be reducible to rational functions.

Geometrically, the construction goes like this: for any point (cos φ, sin φ) on the Unit Circle , draw the line passing through it and the point (−1,0). This point crosses the ''y''-axis at some point ''y'' = ''t''. One can show using simple geometry that ''t'' = tan(φ/2). The equation for the drawn line is ''y'' = (1 + ''x'')''t''. The equation for the intersection of the line and circle is then a Quadratic Equation involving ''t''. The two solutions to this equation are (−1,0) and (cos φ, sin φ). This allows us to write the latter as rational functions of ''t'' (solutions are given below).

:

Note also that the parameter ''t'' represents the Stereographic Projection of the point (cos φ, sin φ) onto the ''y''-axis with the center of projection at (−1,0). Thus, the tangent half-angle formulae give conversions between the stereographic coordinate ''t'' on the unit circle and the standard angular coordinate φ.


IDENTITIES


The tangent half-angle formulae are as follows. Let

:t = an\left( rac{ arphi}{2} ight) = rac{\sin( arphi)}{1 + \cos( arphi)} = rac{1-\cos( arphi)}{\sin( arphi)}.

Then we have
and

Alternatively, from the figure, since
:\overline{DE}=2\sin{ arphi\over2}={2t\over\sqrt{1+t^2}},
one has,
:{\overline{CE}\over\overline{DE}}={\overline{OA}\over\overline{BA}}\,\Rightarrow\,\sin arphi={2t\over1+t^2},
:{\overline{CD}\over\overline{DE}}={\overline{OB}\over\overline{BA}}\,\Rightarrow\,\cos arphi={1-t^2\over1+t^2}.
The other identities follow from those two.

By inverting the exponential formula for ''t'' and finding φ in terms of ''t'', one arrives at the following useful relationship for the Arctangent in terms of the Natural Logarithm
: an^{-1}t = rac{1}{2i}\ln rac{1+it}{1-it}.


USE IN CALCULUS


In Calculus , the tangent half-angle identities can be used in a Trigonometric Substitution to find antiderivatives of rational functions of sin(φ) and cos(φ). After setting

:t= an\left( rac{ arphi}{2} ight),

we have

:dt=\sec^2\left( { arphi \over 2 } ight)\,{ d arphi \over 2}.

Also, since
: arphi=2\arctan(t),
we have
:d arphi = .


HYPERBOLIC IDENTITIES


One can play an entirely analogous game with the Hyperbolic Function s. A point on (the right branch of) a Hyperbola is given by (cosh θ, sinh θ). Projecting this onto ''y''-axis from the center (−1,0) gives the following:

:t = anh\left( rac{ heta}{2} ight) = rac{\sinh( heta)}{\cosh( heta)+1} = rac{\cosh( heta)-1}{\sinh( heta)}

with the identities
and

Finding θ in terms of ''t'' leads to following relationship between the hyperbolic arctangent and the natural logarithm:
: anh^{-1}t = rac{1}{2}\ln rac{1+t}{1-t}


THE GUDERMANNIAN FUNCTION


Comparing the hyperbolic identities to the circular ones, one notices that they involve the same functions of ''t'', just permuted. If we identify the parameter ''t'' in both cases we arrive at a relationship between the circular functions and the hyperbolic ones. That is, if

:t = an rac{ arphi}{2} = anh rac{ heta}{2}

then

: arphi = 2 an^{-1} anh rac{ heta}{2} \equiv \mbox{gd}( heta)
The function gd(θ) is called the Gudermannian Function . The Gudermannian function gives a direct relationship between the circular functions and the hyperbolic ones that does not involve complex numbers. The above descriptions of the tangent half-angle formulae (projection the unit circle and standard hyperbola onto the ''y''-axis) give a nice geometric interpretation of this function.


SEE ALSO