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{{short description|Relates the tangent of half of an angle to trigonometric functions of the entire angle}}
 {{Trigonometry}}In [[Trigonometry|trigonometry]], '''tangent half-angle formulas''' relate the tangent of half of an angle to trigonometric functions of the entire angle. The tangent of half an angle is the [[Stereographic projection|stereographic projection]] of the circle through the point at angle <math display="inline">\pi</math> onto the line through the angles <math display="inline">\pm \frac{\pi}{2}</math>. Among these formulas are the following:

<math display="block">
\begin{align}
\tan \tfrac12( \eta \pm \theta)
&= \frac{\tan \tfrac12 \eta \pm \tan \tfrac12 \theta}{1 \mp \tan \tfrac12 \eta \, \tan \tfrac12 \theta}
= \frac{\sin\eta \pm \sin\theta}{\cos\eta + \cos\theta}
= -\frac{\cos\eta - \cos\theta}{\sin\eta \mp \sin\theta}, \\[10pt]

\tan \tfrac12 \theta
&= \frac{\sin\theta}{1 + \cos\theta}
= \frac{\tan\theta}{\sec\theta + 1}
= \frac{1}{\csc\theta + \cot\theta},
& & (\eta = 0) \\[10pt]

\tan \tfrac12 \theta
&= \frac{1-\cos\theta}{\sin\theta}
= \frac{\sec\theta-1}{\tan\theta}
= \csc\theta-\cot\theta,
& & (\eta = 0) \\[10pt]

\tan \tfrac12 \big(\theta \pm \tfrac12\pi \big)
&= \frac{1 \pm \sin\theta}{\cos\theta}
= \sec\theta \pm \tan\theta
= \frac{\csc\theta \pm 1}{\cot\theta},
& & \big(\eta = \tfrac12\pi \big) \\[10pt]

\tan \tfrac12 \big(\theta \pm \tfrac12\pi \big)
&= \frac{\cos\theta}{1 \mp \sin\theta}
= \frac{1}{\sec\theta \mp \tan\theta}
= \frac{\cot\theta}{\csc\theta \mp 1},
& & \big(\eta = \tfrac12\pi \big) \\[10pt]

\frac{1 - \tan \tfrac12\theta}{1 + \tan \tfrac12\theta}
&= \pm\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} \\[10pt]

\tan \tfrac12 \theta
&= \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} \\[10pt]

\end{align}
</math>

From these one can derive identities expressing the sine, cosine, and tangent as functions of tangents of half-angles:

<math display="block">
\begin{align}
\sin \alpha & = \frac{2\tan \tfrac12 \alpha}{1 + \tan ^2 \tfrac12 \alpha} \\[7pt]
\cos \alpha & = \frac{1 - \tan ^2 \tfrac12 \alpha}{1 + \tan ^2 \tfrac12 \alpha} \\[7pt]
\tan \alpha & = \frac{2\tan \tfrac12 \alpha}{1 - \tan ^2 \tfrac12 \alpha}
\end{align}
</math>

==Proofs==
===Algebraic proofs===
Using double-angle formulae and the Pythagorean identity <math display="inline">1 + \tan^2 \alpha = 1 \big/ \cos^2 \alpha</math> gives

<math display="block">
\sin \alpha
= 2\sin \tfrac12 \alpha \cos \tfrac12 \alpha
= \frac{ 2 \sin \tfrac12 \alpha\, \cos \tfrac12 \alpha
           \Big/ \cos^2 \tfrac12 \alpha}
       {1 + \tan^2 \tfrac12 \alpha}
= \frac{2\tan \tfrac12 \alpha}{1 + \tan^2 \tfrac12 \alpha},
\quad \text{and}
</math>

<math display="block">
\cos \alpha
= \cos^2 \tfrac12 \alpha - \sin^2 \tfrac12 \alpha
= \frac{ \left(\cos^2 \tfrac12 \alpha - \sin^2 \tfrac12 \alpha\right)
           \Big/ \cos^2 \tfrac1 2 \alpha}
       {  1 + \tan^2 \tfrac12 \alpha}
= \frac{1 - \tan^2 \tfrac12 \alpha}{1 + \tan^2 \tfrac12 \alpha}.
</math>


Taking the quotient of the formulae for sine and cosine yields

<math display="block">\tan \alpha = \frac{2\tan \tfrac12 \alpha}{1 - \tan ^2 \tfrac12 \alpha}.</math>

Combining the Pythagorean identity with the double-angle formula for the cosine, <math display="inline"> \cos 2\alpha  =  \cos^2 \alpha - \sin^2 \alpha  =  1 - 2\sin^2 \alpha  =  2\cos^2 \alpha - 1, </math>

rearranging, and taking the square roots yields 

<math display="block"> \left|\sin \alpha\right| = \sqrt {\frac{1-\cos2\alpha}{2}} </math>  and  <math display="block"> \left|\cos \alpha\right| = \sqrt {\frac{1+\cos2\alpha}{2}} </math>

which, upon division gives

<math display="block"> \left|\tan \alpha\right| = \frac {\sqrt {1 - \cos 2\alpha}}{\sqrt {1 + \cos 2\alpha}} = \frac { {\sqrt {1 - \cos 2\alpha}}{\sqrt {1 + \cos 2\alpha}} }{1 + \cos 2\alpha} =\frac{{\sqrt {1 - \cos^2 2\alpha}}}{1 + \cos 2\alpha} = \frac{\left|\sin 2\alpha\right|}{1 + \cos 2\alpha}. </math> 

Alternatively, 

<math display="block"> \left|\tan \alpha\right| = \frac {\sqrt {1 - \cos 2\alpha}}{\sqrt {1 + \cos 2\alpha}} = \frac {1 - \cos 2\alpha}{ {\sqrt {1 + \cos 2\alpha}}{\sqrt {1 - \cos 2\alpha}} } = \frac{1 - \cos 2\alpha}{{\sqrt {1 - \cos^2 2\alpha}}} = \frac{1 - \cos 2\alpha}{\left|\sin 2\alpha\right|}. </math>

It turns out that the absolute value signs in these last two formulas may be dropped, regardless of which quadrant {{mvar|α}} is in.  With or without the absolute value bars these formulas do not apply when both the numerator and denominator on the right-hand side are zero.

Also, using the angle addition and subtraction formulae for both the sine and cosine one obtains:

<math display="block">\begin{align}
  \cos (a+b) &= \cos a \cos b - \sin a \sin b \\
  \cos (a-b) &= \cos a \cos b + \sin a \sin b \\
  \sin (a+b) &= \sin a \cos b + \cos a \sin b \\
  \sin (a-b) &= \sin a \cos b - \cos a \sin b
\end{align}</math>

Pairwise addition of the above four formulae yields:

<math display="block">
\begin{align}
&\sin (a+b) + \sin (a-b) \\[5mu]
&\quad=  \sin a \cos b + \cos a \sin b + \sin a \cos b - \cos a \sin b \\[5mu]
&\quad = 2 \sin a \cos b \\[15mu]

&\cos (a+b) + \cos (a-b) \\[5mu]
&\quad= \cos a \cos b - \sin a \sin b +  \cos a \cos b + \sin a \sin b \\[5mu]
&\quad= 2 \cos a \cos b
\end{align}
</math>

Setting <math display="inline">a= \tfrac12 (p+q)</math> and <math>b= \tfrac12 (p-q)</math> and substituting yields:

<math display="block">
\begin{align}
& \sin p + \sin q \\[5mu]
&\quad= \sin \left(\tfrac12 (p+q) + \tfrac12 (p-q)\right) + \sin\left(\tfrac12(p+q) - \tfrac12 (p-q)\right) \\[5mu]
&\quad= 2 \sin \tfrac12(p+q) \, \cos \tfrac12(p-q) \\[15mu]
& \cos p + \cos q \\[5mu]
&\quad= \cos\left(\tfrac12(p+q) + \tfrac12 (p-q)\right) + \cos\left(\tfrac12(p+q) - \tfrac12(p-q)\right) \\[5mu]
&\quad= 2 \cos\tfrac12(p+q) \, \cos\tfrac12(p-q)
\end{align}
</math>

Dividing the sum of sines by the sum of cosines one arrives at:

<math display="block">
\frac{\sin p + \sin q}{\cos p + \cos q}
= \frac{2 \sin \tfrac12(p+q) \, \cos \tfrac12(p-q)}{2 \cos \tfrac12(p+q) \, \cos \tfrac12(p-q)} = \tan \tfrac12(p+q) </math>

=== Geometric proofs ===
[[File:Tan.half.svg|right|400px|thumb|The sides of this rhombus have length&nbsp;1.  The angle between the horizontal line and the shown diagonal is&nbsp;{{math|{{sfrac|1|2}} (''a'' + ''b'')}}.  This is a geometric way to prove the particular tangent half-angle formula that says {{math|tan {{sfrac|1|2}} (''a'' + ''b'') {{=}} (sin ''a'' + sin ''b'') / (cos ''a'' + cos ''b'')}}. The formulae {{math|sin {{sfrac|1|2}}(''a'' + ''b'')}} and {{math|cos {{sfrac|1|2}}(''a'' + ''b'')}} are the ratios of the actual distances to the length of the diagonal.]]
Applying the formulae derived above to the rhombus figure on the right, it is readily shown that

<math display="block">\tan \tfrac12 (a+b) = \frac{\sin \tfrac12 (a + b)}{\cos \tfrac12 (a + b)} = \frac{\sin a + \sin b}{\cos a + \cos b}.</math>

In the unit circle, application of the above shows that <math display="inline">t = \tan \tfrac12 \varphi</math>. By similarity of triangles,

<math display="block">\frac{t}{\sin \varphi} = \frac{1}{1+ \cos \varphi}.</math>

It follows that

<math display="block">t = \frac{\sin \varphi}{1+ \cos \varphi} = \frac{\sin \varphi(1- \cos \varphi)}{(1+ \cos \varphi)(1- \cos \varphi)} = \frac{1- \cos \varphi}{\sin \varphi}.</math>

{{clear}}

== 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|trigonometry]], it is useful to rewrite the trigonometric functions (such as [[Sine|sine]] and [[Cosine|cosine]]) in terms of [[Rational function|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|calculus]] for converting rational functions in sine and cosine to functions of {{math|''t''}} in order to find their [[Antiderivative|antiderivative]]s.

Geometrically, the construction goes like this: for any point {{math|(cos ''φ'', sin ''φ'')}} on the [[Unit circle|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|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|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>

By eliminating phi between the directly above and the initial definition of <math>t</math>, one arrives at the following useful relationship for the arctangent in terms of the [[Natural logarithm|natural logarithm]]
<math display="block">2 \arctan t = -i \ln\frac{1+it}{1-it}.</math>

In [[Calculus|calculus]], the tangent half-angle substitution substitution is used to find antiderivatives of rational functions of {{math|sin ''φ''}} and&nbsp;{{math|cos ''φ''}}.  After setting

<math display="block">t=\tan\tfrac12\varphi.</math>

This implies that

<math display="block">\varphi=2\arctan(t)+2\pi n , </math>

for some integer {{math|''n''}}, and therefore

<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|hyperbolic function]]s. A point on (the right branch of) a [[Hyperbola|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 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 substitution uses
<math display="block">d\psi = {{2\,dt} \over {1 - t^2}}\,.</math>

==The Gudermannian function==
{{Main|Gudermannian function}}

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

<math display="block">t = \tan\tfrac12 \varphi = \tanh\tfrac12 \psi</math>

then

<math display="block">\varphi = 2\arctan \bigl(\tanh \tfrac12 \psi\,\bigr) \equiv \operatorname{gd} \psi.</math>

where {{math|gd(''ψ'')}} is 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 {{math|''y''}}-axis) give a geometric interpretation of this function.

==Rational values and Pythagorean triples==
{{main article|Pythagorean triple}}
Starting with a [[Pythagorean triangle]] with side lengths {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} that are positive integers and satisfy {{math|''a''{{sup|2}} + ''b''{{sup|2}} {{=}} ''c''{{sup|2}}}}, it follows immediately that each interior angle of the triangle has rational values for sine and cosine, because these are just ratios of side lengths.  Thus each of these angles has a rational value for its half-angle tangent, using {{math|tan ''φ''/2 {{=}} sin ''φ'' / (1 + cos ''φ'')}}.

The reverse is also true.  If there are two positive angles that sum to 90°, each with a rational half-angle tangent, and the third angle is a [[Right angle|right angle]] then a triangle with these interior angles can be scaled to a Pythagorean triangle.  If the third angle is not required to be a right angle, but is the angle that makes the three positive angles sum to 180° then the third angle will necessarily have a rational number for its half-angle tangent when the first two do (using angle addition and subtraction formulas for tangents) and the triangle can be scaled to a [[Heronian triangle]].

Generally, if {{mvar|K}} is a subfield of the complex numbers then {{math|tan ''φ''/2 ∈ ''K'' ∪ {{(}}∞{{)}}}} implies that {{math|{sin ''φ'', cos ''φ'', tan ''φ'', sec ''φ'', csc ''φ'', cot ''φ''} ⊆ ''K'' ∪ {{(}}∞{{)}}}}.

==See also==
*[[List of trigonometric identities]]
*[[Half-side formula]]

==External links==
<!-- * {{springer|title=Tangent formula|id=p/t092150}} Invalid link. -->
* [http://planetmath.org/encyclopedia/TangentOfHalvedAngle.html ''Tangent Of Halved Angle''] at Planetmath
 
{{DEFAULTSORT:Tangent Half-Angle Formula}}
[[Category:Trigonometry]]
[[Category:Conic sections]]
[[Category:Mathematical identities]]

{{Sourceattribution|Tangent half-angle formula}}