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In trigonometry, tangent half-angle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle. Among these are the following

tan⁡(η±θ2)=sin⁡η±sin⁡θcos⁡η+cos⁡θ=−cos⁡η−cos⁡θsin⁡η∓sin⁡θ,tan⁡(±θ2)=±sin⁡θ1+cos⁡θ=±tan⁡θsec⁡θ+1=±1csc⁡θ+cot⁡θ,(η=0)tan⁡(±θ2)=1−cos⁡θ±sin⁡θ=sec⁡θ−1±tan⁡θ=±(csc⁡θ−cot⁡θ),(η=0)tan⁡(12(θ±π2))=1±sin⁡θcos⁡θ=sec⁡θ±tan⁡θ=csc⁡θ±1cot⁡θ,(η=π2)tan⁡(12(θ±π2))=cos⁡θ1∓sin⁡θ=1sec⁡θ∓tan⁡θ=cot⁡θcsc⁡θ∓1,(η=π2)1−tan⁡(θ/2)1+tan⁡(θ/2)=±1−sin⁡θ1+sin⁡θtan⁡θ2=±1−cos⁡θ1+cos⁡θ{displaystyle {begin{aligned}tan left({frac {eta pm theta }{2}}right)&={frac {sin eta pm sin theta }{cos eta +cos theta }}=-{frac {cos eta -cos theta }{sin eta mp sin theta }},\[10pt]tan left(pm {frac {theta }{2}}right)&={frac {pm sin theta }{1+cos theta }}={frac {pm tan theta }{sec theta +1}}={frac {pm 1}{csc theta +cot theta }},&&(eta =0)\[10pt]tan left(pm {frac {theta }{2}}right)&={frac {1-cos theta }{pm sin theta }}={frac {sec theta -1}{pm tan theta }}=pm (csc theta -cot theta ),&&(eta =0)\[10pt]tan left({frac {1}{2}}(theta pm {frac {pi }{2}})right)&={frac {1pm sin theta }{cos theta }}=sec theta pm tan theta ={frac {csc theta pm 1}{cot theta }},&&(eta ={frac {pi }{2}})\[10pt]tan left({frac {1}{2}}(theta pm {frac {pi }{2}})right)&={frac {cos theta }{1mp sin theta }}={frac {1}{sec theta mp tan theta }}={frac {cot theta }{csc theta mp 1}},&&(eta ={frac {pi }{2}})\[10pt]{frac {1-tan(theta /2)}{1+tan(theta /2)}}&=pm {sqrt {frac {1-sin theta }{1+sin theta }}}\[10pt]tan {frac {theta }{2}}&=pm {sqrt {frac {1-cos theta }{1+cos theta }}}end{aligned}}}

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

sin⁡α=2tan⁡α21+tan2⁡α2cos⁡α=1−tan2⁡α21+tan2⁡α2tan⁡α=2tan⁡α21−tan2⁡α2{displaystyle {begin{aligned}sin alpha &={frac {2tan {dfrac {alpha }{2}}}{1+tan ^{2}{dfrac {alpha }{2}}}}\[7pt]cos alpha &={frac {1-tan ^{2}{dfrac {alpha }{2}}}{1+tan ^{2}{dfrac {alpha }{2}}}}\[7pt]tan alpha &={frac {2tan {dfrac {alpha }{2}}}{1-tan ^{2}{dfrac {alpha }{2}}}}end{aligned}}}

Proofs[edit]

Algebraic proofs[edit]

Use double-angle formulae and sin²α + cos²α = 1,

sin⁡α=2sin⁡α2cos⁡α2=2sin⁡α2cos⁡α2cos2⁡α2+sin2⁡α2=2tan⁡α21+tan2⁡α2{displaystyle sin alpha =2sin {frac {alpha }{2}}cos {frac {alpha }{2}}={frac {2sin {frac {alpha }{2}}cos {frac {alpha }{2}}}{cos ^{2}{frac {alpha }{2}}+sin ^{2}{frac {alpha }{2}}}}={frac {2tan {frac {alpha }{2}}}{1+tan ^{2}{frac {alpha }{2}}}}}

cos⁡α=cos2⁡α2−sin2⁡α2=cos2⁡α2−sin2⁡α2cos2⁡α2+sin2⁡α2=cos2⁡α2cos2⁡α2−sin2⁡α2cos2⁡α2cos2⁡α2cos2⁡α2+sin2⁡α2cos2⁡α2=1−tan2⁡α21+tan2⁡α2{displaystyle cos alpha =cos ^{2}{frac {alpha }{2}}-sin ^{2}{frac {alpha }{2}}={frac {cos ^{2}{frac {alpha }{2}}-sin ^{2}{frac {alpha }{2}}}{cos ^{2}{frac {alpha }{2}}+sin ^{2}{frac {alpha }{2}}}}={frac {{frac {cos ^{2}{frac {alpha }{2}}}{cos ^{2}{frac {alpha }{2}}}}-{frac {sin ^{2}{frac {alpha }{2}}}{cos ^{2}{frac {alpha }{2}}}}}{{frac {cos ^{2}{frac {alpha }{2}}}{cos ^{2}{frac {alpha }{2}}}}+{frac {sin ^{2}{frac {alpha }{2}}}{cos ^{2}{frac {alpha }{2}}}}}}={frac {1-tan ^{2}{frac {alpha }{2}}}{1+tan ^{2}{frac {alpha }{2}}}}}

taking the quotient of the formulae for sine and cosine yields

tan⁡α=2tan⁡α21−tan2⁡α2{displaystyle tan alpha ={frac {2tan {frac {alpha }{2}}}{1-tan ^{2}{frac {alpha }{2}}}}}

Combining the Pythagorean identity cos2⁡α+sin2⁡α=1{displaystyle cos ^{2}alpha +sin ^{2}alpha =1} with the double-angle formula for the cosine, cos⁡2α=cos2⁡α−sin2⁡α=1−2sin2⁡α=2cos2⁡α−1{displaystyle cos 2alpha =cos ^{2}alpha -sin ^{2}alpha =1-2sin ^{2}alpha =2cos ^{2}alpha -1},

rearranging, and taking the square roots yields

|sin⁡α|=1−cos⁡2α2{displaystyle |sin alpha |={sqrt {frac {1-cos 2alpha }{2}}}} and |cos⁡α|=1+cos⁡2α2{displaystyle |cos alpha |={sqrt {frac {1+cos 2alpha }{2}}}}

which, upon division gives

|tan⁡α|=1−cos⁡2α1+cos⁡2α{displaystyle |tan alpha |={frac {sqrt {1-cos 2alpha }}{sqrt {1+cos 2alpha }}}} = 1−cos⁡2α1+cos⁡2α1+cos⁡2α{displaystyle {frac {{sqrt {1-cos 2alpha }}{sqrt {1+cos 2alpha }}}{1+cos 2alpha }}} = 1−cos2⁡2α1+cos⁡2α{displaystyle {frac {sqrt {1-cos ^{2}2alpha }}{1+cos 2alpha }}} = |sin⁡2α|1+cos⁡2α{displaystyle {frac {|sin 2alpha |}{1+cos 2alpha }}}

or alternatively

|tan⁡α|=1−cos⁡2α1+cos⁡2α{displaystyle |tan alpha |={frac {sqrt {1-cos 2alpha }}{sqrt {1+cos 2alpha }}}} = 1−cos⁡2α1+cos⁡2α1−cos⁡2α{displaystyle {frac {1-cos 2alpha }{{sqrt {1+cos 2alpha }}{sqrt {1-cos 2alpha }}}}} = 1−cos⁡2α1−cos2⁡2α{displaystyle {frac {1-cos 2alpha }{sqrt {1-cos ^{2}2alpha }}}} = 1−cos⁡2α|sin⁡2α|{displaystyle {frac {1-cos 2alpha }{|sin 2alpha |}}}.

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

cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b{displaystyle cos(a+b)=cos acos b-sin asin b}

cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b{displaystyle cos(a-b)=cos acos b+sin asin b}

sin⁡(a+b)=sin⁡acos⁡b+cos⁡asin⁡b{displaystyle sin(a+b)=sin acos b+cos asin b}

sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b{displaystyle sin(a-b)=sin acos b-cos asin b}

Pairwise addition of the above four formulae yields:

sin⁡(a+b)+sin⁡(a−b)=sin⁡acos⁡b+cos⁡asin⁡b+sin⁡acos⁡b−cos⁡asin⁡b=2sin⁡acos⁡bcos⁡(a+b)+cos⁡(a−b)=cos⁡acos⁡b−sin⁡asin⁡b+cos⁡acos⁡b+sin⁡asin⁡b=2cos⁡acos⁡b{displaystyle {begin{aligned}sin(a+b)+sin(a-b)&=sin acos b+cos asin b+sin acos b-cos asin b\&=2sin acos b\[3pt]cos(a+b)+cos(a-b)&=cos acos b-sin asin b+cos acos b+sin asin b\&=2cos acos bend{aligned}}}

Setting a=p+q2{displaystyle a={frac {p+q}{2}}} and b=p−q2{displaystyle b={frac {p-q}{2}}} and substituting yields:

sin⁡(p+q2+p−q2)+sin⁡(p+q2−p−q2)=sin⁡(p)+sin⁡(q)=2sin⁡(p+q2)cos⁡(p−q2)cos⁡(p+q2+p−q2)+cos⁡(p+q2−p−q2)=cos⁡(p)+cos⁡(q)=2cos⁡(p+q2)cos⁡(p−q2){displaystyle {begin{aligned}sin left({frac {p+q}{2}}+{frac {p-q}{2}}right)+sin left({frac {p+q}{2}}-{frac {p-q}{2}}right)&=sin(p)+sin(q)\&=2sin left({frac {p+q}{2}}right)cos left({frac {p-q}{2}}right)\[6pt]cos left({frac {p+q}{2}}+{frac {p-q}{2}}right)+cos left({frac {p+q}{2}}-{frac {p-q}{2}}right)&=cos(p)+cos(q)\&=2cos left({frac {p+q}{2}}right)cos left({frac {p-q}{2}}right)end{aligned}}}

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

sin⁡(p)+sin⁡(q)cos⁡(p)+cos⁡(q)=2sin⁡(p+q2)cos⁡(p−q2)2cos⁡(p+q2)cos⁡(p−q2)=tan⁡(p+q2){displaystyle {frac {sin(p)+sin(q)}{cos(p)+cos(q)}}={frac {2sin left({frac {p+q}{2}}right)cos left({frac {p-q}{2}}right)}{2cos left({frac {p+q}{2}}right)cos left({frac {p-q}{2}}right)}}=tan left({frac {p+q}{2}}right)}

Geometric proofs[edit]

Applying the formulae derived above to the rhombus figure on the right, it is readily shown that

The sides of this rhombus have length 1. The angle between the horizontal line and the shown diagonal is (a + b)/2. This is a geometric way to prove a tangent half-angle formula. Note that sin((a + b)/2) and cos((a + b)/2) just show their relation to the diagonal, not the real value.

tan⁡a+b2=sin⁡a+b2cos⁡a+b2=sin⁡a+sin⁡bcos⁡a+cos⁡b.{displaystyle tan {frac {a+b}{2}}={frac {sin {frac {a+b}{2}}}{cos {frac {a+b}{2}}}}={frac {sin a+sin b}{cos a+cos b}}.}

In the unit circle, application of the above shows that t=tan⁡(φ2){displaystyle t=tan left({frac {varphi }{2}}right)}. According to similar triangles,

A geometric proof of the tangent half-angle formula

tsin⁡φ=11+cos⁡φ{displaystyle {frac {t}{sin varphi }}={frac {1}{1+cos varphi }}}. It follows that t=sin⁡φ1+cos⁡φ=sin⁡φ(1−cos⁡φ)(1+cos⁡φ)(1−cos⁡φ)=1−cos⁡φsin⁡φ.{displaystyle 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 }}.}

The tangent half-angle substitution in integral calculus[edit]

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 t. These identities are known collectively as the tangent half-angle formulae because of the definition of t. These identities can be useful in calculus for converting rational functions in sine and cosine to functions of t in order to find their antiderivatives.

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.

Then we have

cos⁡φ=1−t21+t2,sin⁡φ=2t1+t2,tan⁡φ=2t1−t2cot⁡φ=1−t22t,sec⁡φ=1+t21−t2,csc⁡φ=1+t22t,{displaystyle {begin{aligned}&cos varphi ={frac {1-t^{2}}{1+t^{2}}},&&sin varphi ={frac {2t}{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{aligned}}}

and

eiφ=1+it1−it,e−iφ=1−it1+it.{displaystyle e^{ivarphi }={frac {1+it}{1-it}},qquad e^{-ivarphi }={frac {1-it}{1+it}}.}

By eliminating phi between the directly above and the initial definition of t, one arrives at the following useful relationship for the arctangent in terms of the natural logarithm

arctan⁡t=12iln⁡1+it1−it.{displaystyle arctan t={frac {1}{2i}}ln {frac {1+it}{1-it}}.}

In calculus, the Weierstrass substitution is used to find antiderivatives of rational functions of sin φ and cos φ. After setting

t=tan⁡12φ.{displaystyle t=tan {tfrac {1}{2}}varphi .}

This implies that

φ=2arctan⁡(t)+2πn,{displaystyle varphi =2arctan(t)+2pi n,}

for some integer n, and therefore

dφ=2dt1+t2.{displaystyle dvarphi ={{2,dt} over {1+t^{2}}}.}

Hyperbolic identities[edit]

One can play an entirely analogous game with the hyperbolic functions. 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=tanh⁡12θ=sinh⁡θcosh⁡θ+1=cosh⁡θ−1sinh⁡θ{displaystyle t=tanh {tfrac {1}{2}}theta ={frac {sinh theta }{cosh theta +1}}={frac {cosh theta -1}{sinh theta }}}

with the identities

cosh⁡θ=1+t21−t2,sinh⁡θ=2t1−t2,tanh⁡θ=2t1+t2,coth⁡θ=1+t22t,sechθ=1−t21+t2,cschθ=1−t22t,{displaystyle {begin{aligned}&cosh theta ={frac {1+t^{2}}{1-t^{2}}},&&sinh theta ={frac {2t}{1-t^{2}}},\[8pt]&tanh theta ={frac {2t}{1+t^{2}}},&&coth theta ={frac {1+t^{2}}{2t}},\[8pt]&operatorname {sech} ,theta ={frac {1-t^{2}}{1+t^{2}}},&&operatorname {csch} ,theta ={frac {1-t^{2}}{2t}},end{aligned}}}

and

eθ=1+t1−t,e−θ=1−t1+t.{displaystyle e^{theta }={frac {1+t}{1-t}},qquad e^{-theta }={frac {1-t}{1+t}}.}

The use of this substitution for finding antiderivatives was introduced by Karl Weierstrass.^{[citation needed]}

Finding θ in terms of t leads to following relationship between the hyperbolic arctangent and the natural logarithm:

artanh⁡t=12ln⁡1+t1−t.{displaystyle operatorname {artanh} t={frac {1}{2}}ln {frac {1+t}{1-t}}.}

("ar-" is used rather than "arc-" because "arc" is about arc length and "ar" abbreviates "area". It is the area between two rays and a hyperbola, rather than the arc length between two rays measure along an arc a circle.)

The Gudermannian function[edit]

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=tan⁡12φ=tanh⁡12θ{displaystyle t=tan {tfrac {1}{2}}varphi =tanh {tfrac {1}{2}}theta }

then

φ=2tan−1⁡tanh⁡12θ≡gd⁡θ.{displaystyle varphi =2tan ^{-1}tanh {tfrac {1}{2}}theta equiv operatorname {gd} theta .}

where gd(θ) 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 y-axis) give a geometric interpretation of this function.

Pythagorean triples[edit]

The tangent of half of an acute angle of a right triangle whose sides are a Pythagorean triple will necessarily be a rational number in the interval (0, 1). Vice versa, when a half-angle tangent is a rational number in the interval (0, 1), there is a right triangle that has the full angle and that has side lengths that are a Pythagorean triple.

See also[edit]

• List of trigonometric identities


• Half-side formula

External links[edit]

• 
  Tangent Of Halved Angle at Planetmath