Template:Short description Template:TrigonometryIn trigonometry, tangent half-angle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle.<ref>Mathematics. United States, NAVEDTRA [i.e. Naval] Education and Training Program Management Support Activity, 1989. 6-19.</ref>
FormulaeEdit
The tangent of half an angle is the stereographic projection of the circle through the point at angle <math display="inline">\pi</math> radians onto the line through the angles <math display="inline">\pm \frac{\pi}{2}</math>. Tangent half-angle formulae include <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}\,, \end{align} </math> with simpler formulae when Template:Mvar is known to be Template:Math, Template:Math, Template:Math, or Template:Math because Template:Math and Template:Math can be replaced by simple constants.
In the reverse direction, the formulae include <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>
ProofsEdit
Algebraic proofsEdit
Using the angle addition and subtraction formulae for both the sine and cosine one obtains <math display="block"> \begin{align} \sin (a+b) + \sin (a-b) &= 2 \sin a \cos b \\[15mu] \cos (a+b) + \cos (a-b) & = 2 \cos a \cos b\,. \end{align} </math>
Setting <math display="inline">a= \tfrac12 (\eta+\theta)</math> and <math>b= \tfrac12 (\eta-\theta)</math> and substituting yields <math display="block"> \begin{align} \sin \eta + \sin \theta = 2 \sin \tfrac12(\eta+\theta) \, \cos \tfrac12(\eta-\theta) \\[15mu] \cos \eta + \cos \theta = 2 \cos\tfrac12(\eta+\theta) \, \cos\tfrac12(\eta-\theta)\,. \end{align} </math>
Dividing the sum of sines by the sum of cosines gives <math display="block">\frac{\sin \eta + \sin \theta}{\cos \eta + \cos \theta} = \tan \tfrac12(\eta+\theta)\,.</math>
Also, a similar calculation starting with <math>\sin (a+b) - \sin (a-b)</math> and <math>\cos (a+b) - \cos (a-b)</math> gives <math display="block">-\frac{\cos \eta - \cos \theta}{\sin \eta - \sin \theta} = \tan \tfrac12(\eta+\theta)\,.</math>
Furthermore, 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} </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>
Geometric proofsEdit
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>
The tangent half-angle substitution in integral calculusEdit
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In various applications of trigonometry, it is useful to rewrite the trigonometric functions (such as sine and cosine) in terms of rational functions 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 Template:Math in order to find their antiderivatives.
Geometrically, the construction goes like this: for any point Template:Math on the unit circle, draw the line passing through it and the point Template:Math. This point crosses the Template:Math-axis at some point Template:Math. One can show using simple geometry that Template:Math. The equation for the drawn line is Template:Math. The equation for the intersection of the line and circle is then a quadratic equation involving Template:Math. The two solutions to this equation are Template:Math and Template:Math. This allows us to write the latter as rational functions of Template:Math (solutions are given below).
The parameter Template:Math represents the stereographic projection of the point Template:Math onto the Template:Math-axis with the center of projection at Template:Math. Thus, the tangent half-angle formulae give conversions between the stereographic coordinate Template:Math on the unit circle and the standard angular coordinate Template:Math.
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 Template:Math and Template:Math. 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 identitiesEdit
One can play an entirely analogous game with the hyperbolic functions. A point on (the right branch of) a hyperbola is given by Template:Math. Projecting this onto Template:Math-axis from the center Template:Math 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 Template:Math in terms of Template:Math 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 in calculus uses <math display="block">d\psi = {{2\,dt} \over {1 - t^2}}\,.</math>
The Gudermannian functionEdit
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Comparing the hyperbolic identities to the circular ones, one notices that they involve the same functions of Template:Math, just permuted. If we identify the parameter Template:Math 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 Template:Math 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 Template:Math-axis) give a geometric interpretation of this function.
Rational values and Pythagorean triplesEdit
Template:Main article Starting with a Pythagorean triangle with side lengths Template:Mvar, Template:Mvar, and Template:Mvar that are positive integers and satisfy Template:Math, 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 Template:Math.
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 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 Template:Mvar is a subfield of the complex numbers then Template:Math implies that Template:Math.