Editing Tangent half-angle formula (section)

== The tangent half-angle substitution in integral calculus ==

{{Main|Tangent half-angle substitution}}

[[Image:Weierstrass substitution.svg|right|400px|thumb|A [[geometric]] proof of the tangent half-angle substitution]]

In various applications of [[trigonometry]], it is useful to rewrite the [[trigonometric function]]s (such as [[sine]] and [[cosine]]) in terms of [[rational function]]s of a new variable <math>t</math>. These identities are known collectively as the '''tangent half-angle formulae''' because of the definition of <math>t</math>. These identities can be useful in [[calculus]] for converting rational functions in sine and cosine to functions of {{math|''t''}} in order to find their [[antiderivative]]s.

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

The parameter {{math|''t''}} represents the [[stereographic projection]] of the point {{math|(cos ''φ'', sin ''φ'')}} onto the {{math|''y''}}-axis with the center of projection at {{math|(−1, 0)}}. Thus, the tangent half-angle formulae give conversions between the stereographic coordinate {{math|''t''}} on the unit circle and the standard angular coordinate {{math|''φ''}}.

Then we have

<math display="block">
\begin{align}
& \sin\varphi = \frac{2t}{1 + t^2},
& & \cos\varphi = \frac{1 - t^2}{1 + t^2}, \\[8pt]
& \tan\varphi = \frac{2t}{1 - t^2}
& & \cot\varphi = \frac{1 - t^2}{2t}, \\[8pt]
& \sec\varphi = \frac{1 + t^2}{1 - t^2},
& & \csc\varphi = \frac{1 + t^2}{2t},
\end{align}
</math>

and

<math display="block">e^{i \varphi} = \frac{1 + i t}{1 - i t}, \qquad
e^{-i \varphi} = \frac{1 - i t}{1 + i t}.
</math>

Both this expression of <math>e^{i\varphi}</math> and the expression <math>t = \tan(\varphi/2)</math> can be solved for <math>\varphi</math>.  Equating these gives the [[arctangent]] in terms of the [[natural logarithm]]
<math display="block">\arctan t = \frac{-i}{2} \ln\frac{1+it}{1-it}.</math>

In [[calculus]], the tangent half-angle substitution is used to find antiderivatives of [[rational functions]] of {{math|sin ''φ''}} and&nbsp;{{math|cos ''φ''}}.  Differentiating <math>t=\tan\tfrac12\varphi</math> gives 
<math display="block">\frac{dt}{d\varphi} = \tfrac12\sec^2 \tfrac12\varphi = \tfrac12(1+\tan^2 \tfrac12\varphi) = \tfrac12(1+t^2)</math>
and thus
<math display="block">d\varphi = {{2\,dt} \over {1 + t^2}}.</math>

===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&nbsp;{{math|(cosh ''ψ'', sinh ''ψ'')}}. Projecting this onto {{math|''y''}}-axis from the center {{math|(−1, 0)}} gives the following:

<math display="block">t = \tanh\tfrac12\psi = \frac{\sinh\psi}{\cosh\psi+1} = \frac{\cosh\psi-1}{\sinh\psi}</math>

with the identities

<math display="block">
\begin{align}
& \sinh\psi = \frac{2t}{1 - t^2},
& & \cosh\psi = \frac{1 + t^2}{1 - t^2}, \\[8pt]
& \tanh\psi = \frac{2t}{1 + t^2},
& & \coth\psi = \frac{1 + t^2}{2t}, \\[8pt]
& \operatorname{sech}\,\psi = \frac{1 - t^2}{1 + t^2},
& & \operatorname{csch}\,\psi = \frac{1 - t^2}{2t},
\end{align}
</math>

and

<math display="block">e^\psi = \frac{1 + t}{1 - t}, \qquad
e^{-\psi} = \frac{1 - t}{1 + t}.</math>

Finding {{math|''ψ''}} in terms of {{math|''t''}} leads to following relationship between the [[inverse hyperbolic functions|inverse hyperbolic tangent]] <math>\operatorname{artanh}</math> and the natural logarithm:

<math display="block">2 \operatorname{artanh} t = \ln\frac{1+t}{1-t}.</math>

The hyperbolic tangent half-angle substitution in calculus uses
<math display="block">d\psi = {{2\,dt} \over {1 - t^2}}\,.</math>