In trigonometry, tangent half-angle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle.
Formulae The tangent of half an angle is the stereographic projection of the circle through the point at angle π {\textstyle \pi } radians onto the line through the angles ± π 2 {\textstyle \pm {\frac {\pi }{2}}} . Tangent half-angle formulae include
tan 1 2 ( η ± θ ) = tan 1 2 η ± tan 1 2 θ 1 ∓ tan 1 2 η tan 1 2 θ = sin η ± sin θ cos η + cos θ = − cos η − cos θ sin η ∓ sin θ , {\displaystyle {\begin{aligned}\tan {\tfrac {1}{2}}(\eta \pm \theta )&={\frac {\tan {\tfrac {1}{2}}\eta \pm \tan {\tfrac {1}{2}}\theta }{1\mp \tan {\tfrac {1}{2}}\eta \,\tan {\tfrac {1}{2}}\theta }}={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}=-{\frac {\cos \eta -\cos \theta }{\sin \eta \mp \sin \theta }}\,,\end{aligned}}}
with simpler formulae when η is known to be 0, π/2, π, or 3π/2 because sin(η) and cos(η) can be replaced by simple constants. In the reverse direction, the formulae include
sin α = 2 tan 1 2 α 1 + tan 2 1 2 α cos α = 1 − tan 2 1 2 α 1 + tan 2 1 2 α tan α = 2 tan 1 2 α 1 − tan 2 1 2 α . {\displaystyle {\begin{aligned}\sin \alpha &={\frac {2\tan {\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}\\[7pt]\cos \alpha &={\frac {1-\tan ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}\\[7pt]\tan \alpha &={\frac {2\tan {\tfrac {1}{2}}\alpha }{1-\tan ^{2}{\tfrac {1}{2}}\alpha }}\,.\end{aligned}}}
Proofs
Algebraic proofs Using the angle addition and subtraction formulae for both the sine and cosine one obtains
sin ( a + b ) + sin ( a − b ) = 2 sin a cos b cos ( a + b ) + cos ( a − b ) = 2 cos a cos b . {\displaystyle {\begin{aligned}\sin(a+b)+\sin(a-b)&=2\sin a\cos b\\[15mu]\cos(a+b)+\cos(a-b)&=2\cos a\cos b\,.\end{aligned}}}
Setting a = 1 2 ( η + θ ) {\textstyle a={\tfrac {1}{2}}(\eta +\theta )} and b = 1 2 ( η − θ ) {\displaystyle b={\tfrac {1}{2}}(\eta -\theta )} and substituting yields
sin η + sin θ = 2 sin 1 2 ( η + θ ) cos 1 2 ( η − θ ) cos η + cos θ = 2 cos 1 2 ( η + θ ) cos 1 2 ( η − θ ) . {\displaystyle {\begin{aligned}\sin \eta +\sin \theta &=2\sin {\tfrac {1}{2}}(\eta +\theta )\,\cos {\tfrac {1}{2}}(\eta -\theta )\\[15mu]\cos \eta +\cos \theta &=2\cos {\tfrac {1}{2}}(\eta +\theta )\,\cos {\tfrac {1}{2}}(\eta -\theta )\,.\end{aligned}}}
Dividing the sum of sines by the sum of cosines gives
sin η + sin θ cos η + cos θ = tan 1 2 ( η + θ ) . {\displaystyle {\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}=\tan {\tfrac {1}{2}}(\eta +\theta )\,.}
Also, a similar calculation starting with sin ( a + b ) − sin ( a − b ) {\displaystyle \sin(a+b)-\sin(a-b)} and cos ( a + b ) − cos ( a − b ) {\displaystyle \cos(a+b)-\cos(a-b)} gives
− cos η − cos θ sin η − sin θ = tan 1 2 ( η + θ ) . {\displaystyle -{\frac {\cos \eta -\cos \theta }{\sin \eta -\sin \theta }}=\tan {\tfrac {1}{2}}(\eta +\theta )\,.}
Furthermore, using double-angle formulae and the Pythagorean identity 1 + tan 2 1 2 α = 1 / cos 2 1 2 α {\textstyle 1+\tan ^{2}{\tfrac {1}{2}}\alpha =1{\big /}\cos ^{2}{\tfrac {1}{2}}\alpha } gives
sin α = 2 sin 1 2 α cos 1 2 α = 2 sin 1 2 α cos 1 2 α / cos 2 1 2 α 1 + tan 2 1 2 α = 2 tan 1 2 α 1 + tan 2 1 2 α {\displaystyle \sin \alpha =2\sin {\tfrac {1}{2}}\alpha \cos {\tfrac {1}{2}}\alpha ={\frac {2\sin {\tfrac {1}{2}}\alpha \,\cos {\tfrac {1}{2}}\alpha {\Big /}\cos ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}={\frac {2\tan {\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}}
cos α = cos 2 1 2 α − sin 2 1 2 α = ( cos 2 1 2 α − sin 2 1 2 α ) / cos 2 1 2 α 1 + tan 2 1 2 α = 1 − tan 2 1 2 α 1 + tan 2 1 2 α . {\displaystyle \cos \alpha =\cos ^{2}{\tfrac {1}{2}}\alpha -\sin ^{2}{\tfrac {1}{2}}\alpha ={\frac {\left(\cos ^{2}{\tfrac {1}{2}}\alpha -\sin ^{2}{\tfrac {1}{2}}\alpha \right){\Big /}\cos ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}={\frac {1-\tan ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}\,.}
Taking the quotient of the formulae for sine and cosine yields
tan α = 2 tan 1 2 α 1 − tan 2 1 2 α . {\displaystyle \tan \alpha ={\frac {2\tan {\tfrac {1}{2}}\alpha }{1-\tan ^{2}{\tfrac {1}{2}}\alpha }}\,.}
Geometric proofs
Applying the formulae derived above to the rhombus figure on the right, it is readily shown that
tan 1 2 ( a + b ) = sin 1 2 ( a + b ) cos 1 2 ( a + b ) = sin a + sin b cos a + cos b . {\displaystyle \tan {\tfrac {1}{2}}(a+b)={\frac {\sin {\tfrac {1}{2}}(a+b)}{\cos {\tfrac {1}{2}}(a+b)}}={\frac {\sin a+\sin b}{\cos a+\cos b}}.}
In the unit circle, application of the above shows that t = tan 1 2 φ {\textstyle t=\tan {\tfrac {1}{2}}\varphi } . By similarity of triangles,
t sin φ = 1 1 + cos φ . {\displaystyle {\frac {t}{\sin \varphi }}={\frac {1}{1+\cos \varphi }}.}
It follows that
t = sin φ 1 + cos φ = sin φ ( 1 − cos φ ) ( 1 + cos φ ) ( 1 − cos φ ) = 1 − cos φ sin 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
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 {\displaystyle t} . These identities are known collectively as the tangent half-angle formulae because of the definition of t {\displaystyle 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. 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). 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
sin φ = 2 t 1 + t 2 , cos φ = 1 − t 2 1 + t 2 , tan φ = 2 t 1 − t 2 cot φ = 1 − t 2 2 t , sec φ = 1 + t 2 1 − t 2 , csc φ = 1 + t 2 2 t , {\displaystyle {\begin{aligned}&\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{aligned}}}
and
e i φ = 1 + i t 1 − i t , e − i φ = 1 − i t 1 + i t . {\displaystyle e^{i\varphi }={\frac {1+it}{1-it}},\qquad e^{-i\varphi }={\frac {1-it}{1+it}}.}
Both this expression of e i φ {\displaystyle e^{i\varphi }} and the expression t = tan ( φ / 2 ) {\displaystyle t=\tan(\varphi /2)} can be solved for φ {\displaystyle \varphi } . Equating these gives the arctangent in terms of the natural logarithm
arctan t = − i 2 ln 1 + i t 1 − i t . {\displaystyle \arctan t={\frac {-i}{2}}\ln {\frac {1+it}{1-it}}.}
In calculus, the tangent half-angle substitution is used to find antiderivatives of rational functions of sin φ and cos φ. Differentiating t = tan 1 2 φ {\displaystyle t=\tan {\tfrac {1}{2}}\varphi } gives
d t d φ = 1 2 sec 2 1 2 φ = 1 2 ( 1 + tan 2 1 2 φ ) = 1 2 ( 1 + t 2 ) {\displaystyle {\frac {dt}{d\varphi }}={\tfrac {1}{2}}\sec ^{2}{\tfrac {1}{2}}\varphi ={\tfrac {1}{2}}(1+\tan ^{2}{\tfrac {1}{2}}\varphi )={\tfrac {1}{2}}(1+t^{2})}
and thus
d φ = 2 d t 1 + t 2 . {\displaystyle d\varphi ={{2\,dt} \over {1+t^{2}}}.}
Hyperbolic identities One can play an entirely analogous game with the hyperbolic functions. A point on (the right branch of) a hyperbola is given by (cosh ψ, sinh ψ). Projecting this onto y-axis from the center (−1, 0) gives the following:
t = tanh 1 2 ψ = sinh ψ cosh ψ + 1 = cosh ψ − 1 sinh ψ {\displaystyle t=\tanh {\tfrac {1}{2}}\psi ={\frac {\sinh \psi }{\cosh \psi +1}}={\frac {\cosh \psi -1}{\sinh \psi }}}
with the identities
sinh ψ = 2 t 1 − t 2 , cosh ψ = 1 + t 2 1 − t 2 , tanh ψ = 2 t 1 + t 2 , coth ψ = 1 + t 2 2 t , sech ψ = 1 − t 2 1 + t 2 , csch ψ = 1 − t 2 2 t , {\displaystyle {\begin{aligned}&\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{aligned}}}
and
e ψ = 1 + t 1 − t , e − ψ = 1 − t 1 + t . {\displaystyle e^{\psi }={\frac {1+t}{1-t}},\qquad e^{-\psi }={\frac {1-t}{1+t}}.}
Finding ψ in terms of t leads to following relationship between the inverse hyperbolic tangent artanh {\displaystyle \operatorname {artanh} } and the natural logarithm:
2 artanh t = ln 1 + t 1 − t . {\displaystyle 2\operatorn