Tangent half-angle formula

In trigonometry, tangent half-angle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle.^[1]

Formulae

The tangent of half an angle is the stereographic projection of the circle through the point at angle \pi radians onto the line through the angles \pm \frac{\pi}{2}. Tangent half-angle formulae include

\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}

with simpler formulae when η is known to be $0$ , $π /2$ , $π$ , or $3 π /2$ because $sin(η)$ and $cos(η)$ can be replaced by simple constants.

In the reverse direction, the formulae include

\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}

Proofs

Algebraic proofs

Using the angle addition and subtraction formulae for both the sine and cosine one obtains

\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}

Setting a= \tfrac12 (\eta+\theta) and b= \tfrac12 (\eta-\theta) and substituting yields

\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}

Dividing the sum of sines by the sum of cosines gives

\frac{\sin \eta + \sin \theta}{\cos \eta + \cos \theta} = \tan \tfrac12(\eta+\theta)\,.

Also, a similar calculation starting with \sin (a+b) - \sin (a-b) and \cos (a+b) - \cos (a-b) gives

-\frac{\cos \eta - \cos \theta}{\sin \eta - \sin \theta} = \tan \tfrac12(\eta+\theta)\,.

Furthermore, using double-angle formulae and the Pythagorean identity 1 + \tan^2 \tfrac12 \alpha = 1 \big/ \cos^2 \tfrac12 \alpha gives

\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}

\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}\,.

Taking the quotient of the formulae for sine and cosine yields

\tan \alpha = \frac{2\tan \tfrac12 \alpha}{1 - \tan ^2 \tfrac12 \alpha}\,.

Geometric proofs

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

\tan \tfrac12 (a+b) = \frac{\sin \tfrac12 (a + b)}{\cos \tfrac12 (a + b)} = \frac{\sin a + \sin b}{\cos a + \cos b}.

In the unit circle, application of the above shows that t = \tan \tfrac12 \varphi. By similarity of triangles,

\frac{t}{\sin \varphi} = \frac{1}{1+ \cos \varphi}.

It follows that

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. 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.

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

\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}

and

e^{i \varphi} = \frac{1 + i t}{1 - i t}, \qquad
e^{-i \varphi} = \frac{1 - i t}{1 + i t}.

Both this expression of e^{i\varphi} and the expression t = \tan(\varphi/2) can be solved for \varphi. Equating these gives the arctangent in terms of the natural logarithm

\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\tfrac12\varphi gives

\frac{dt}{d\varphi} = \tfrac12\sec^2 \tfrac12\varphi = \tfrac12(1+\tan^2 \tfrac12\varphi) = \tfrac12(1+t^2)

and thus

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\tfrac12\psi = \frac{\sinh\psi}{\cosh\psi+1} = \frac{\cosh\psi-1}{\sinh\psi}

with the identities

\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}

and

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 \operatorname{artanh} and the natural logarithm:

2 \operatorname{artanh} t = \ln\frac{1+t}{1-t}.

The hyperbolic tangent half-angle substitution in calculus uses

d\psi = {{2\,dt} \over {1 - t^2}}\,.

The Gudermannian function

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\tfrac12 \varphi = \tanh\tfrac12 \psi

then

\varphi = 2\arctan \bigl(\tanh \tfrac12 \psi\,\bigr) \equiv \operatorname{gd} \psi.

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

Rational values and Pythagorean triples

Starting with a Pythagorean triangle with side lengths a, b, and c that are positive integers and satisfy $a 2 + b 2 = c 2$ , 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 $tan φ /2 = sin φ / (1 + cos φ)$ .

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 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 K is a subfield of the complex numbers then $tan φ /2 \in K \cup$ implies that ${sin φ, cos φ, tan φ, sec φ, csc φ, cot φ} \subseteq K \cup$ .

References

^ Mathematics. United States, NAVEDTRA [i.e. Naval] Education and Training Program Management Support Activity, 1989. 6-19.