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. The tangent of half an angle is the stereographic projection of the circle onto a line. Among these formulas are the following:

:$\begin \tan \tfrac12( \eta \pm \theta) &= \frac = \frac = -\frac, \\$0pt
\tan \tfrac12 \theta
&= \frac
= \frac
= \frac,
& & (\eta = 0) \\0pt
\tan \tfrac12 \theta
&= \frac
= \frac
= \csc\theta-\cot\theta,
& & (\eta = 0) \\0pt
\tan \tfrac12 \big(\theta \pm \tfrac12\pi \big)
&= \frac
= \sec\theta \pm \tan\theta
= \frac,
& & \big(\eta = \tfrac12\pi \big) \\0pt
\tan \tfrac12 \big(\theta \pm \tfrac12\pi \big)
&= \frac
= \frac
= \frac,
& & \big(\eta = \tfrac12\pi \big) \\0pt
\frac
&= \pm\sqrt \\0pt
\tan \tfrac12 \theta
&= \pm \sqrt \\0pt
\end

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

:
\begin
\sin \alpha & = \frac \\ pt\cos \alpha & = \frac \\ pt\tan \alpha & = \frac
\end

Proofs

Algebraic proofs

Using double-angle formulae and the Pythagorean identity $1 + \tan^2 \alpha = 1 \big/ \cos^2 \alpha$ gives

: $\sin \alpha = 2\sin \tfrac12 \alpha \cos \tfrac12 \alpha = \frac = \frac, \quad \text$

: $\cos \alpha = \cos^2 \tfrac12 \alpha - \sin^2 \tfrac12 \alpha = \frac = \frac, \quad \text$

Taking the quotient of the formulae for sine and cosine yields

: $\tan \alpha = \frac.$

Combining the Pythagorean identity with the double-angle formula for the cosine, $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = 1 - 2\sin^2 \alpha = 2\cos^2 \alpha - 1$,

rearranging, and taking the square roots yields

: $, \sin \alpha, = \sqrt$ and $, \cos \alpha, = \sqrt$

which, upon division gives

: $, \tan \alpha, = \frac = \frac =\frac = \frac.$

Alternatively,

:$, \tan \alpha, = \frac = \frac = \frac = \frac.$

It turns out that the absolute value signs in these last two formulas may be dropped, regardless of which quadrant 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:

: $\cos (a+b) = \cos a \cos b - \sin a \sin b$

: $\cos (a-b) = \cos a \cos b + \sin a \sin b$

: $\sin (a+b) = \sin a \cos b + \cos a \sin b$

: $\sin (a-b) = \sin a \cos b - \cos a \sin b$

Pairwise addition of the above four formulae yields:

:
\begin
&\sin (a+b) + \sin (a-b) \\ mu&\quad= \sin a \cos b + \cos a \sin b + \sin a \cos b - \cos a \sin b \\ mu&\quad = 2 \sin a \cos b \\ 5mu
&\cos (a+b) + \cos (a-b) \\ mu&\quad= \cos a \cos b - \sin a \sin b + \cos a \cos b + \sin a \sin b \\ mu&\quad= 2 \cos a \cos b
\end

Setting $a= \tfrac12 (p+q)$ and $b= \tfrac12 (p-q)$ and substituting yields:

:
\begin
& \sin p + \sin q \\ mu&\quad= \sin \left(\tfrac12 (p+q) + \tfrac12 (p-q)\right) + \sin\left(\tfrac12(p+q) - \tfrac12 (p-q)\right) \\ mu&\quad= 2 \sin \tfrac12(p+q) \, \cos \tfrac12(p-q) \\ 5mu& \cos p + \cos q \\ mu&\quad= \cos\left(\tfrac12(p+q) + \tfrac12 (p-q)\right) + \cos\left(\tfrac12(p+q) - \tfrac12(p-q)\right) \\ mu&\quad= 2 \cos\tfrac12(p+q) \, \cos\tfrac12(p-q)
\end

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

: $\frac = \frac = \tan \tfrac12(p+q)$

Geometric proofs

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

: $\tan \tfrac12 (a+b) = \frac = \frac.$

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

: $\frac = \frac$.

It follows that

: $t = \frac = \frac = \frac.$

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 in order to find their antiderivatives.

Geometrically, the construction goes like this: for any point on the unit circle, draw the line passing through it and the point . This point crosses the -axis at some point . One can show using simple geometry that . The equation for the drawn line is . The equation for the intersection of the line and circle is then a quadratic equation involving . The two solutions to this equation are and . This allows us to write the latter as rational functions of (solutions are given below).

The parameter represents the stereographic projection of the point onto the -axis with the center of projection at . Thus, the tangent half-angle formulae give conversions between the stereographic coordinate on the unit circle and the standard angular coordinate .

Then we have

:
\begin
& \sin\varphi = \frac,
& & \cos\varphi = \frac, \\ pt& \tan\varphi = \frac
& & \cot\varphi = \frac, \\ pt& \sec\varphi = \frac,
& & \csc\varphi = \frac,
\end

and

: $e^ = \frac, \qquad e^ = \frac.$

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
:$2 \arctan t = -i \ln\frac.$

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

:$t=\tan\tfrac12\varphi.$

This implies that

:$\varphi=2\arctan(t)+2\pi n ,$

for some integer , and therefore

:$d\varphi = .$

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 . Projecting this onto -axis from the center gives the following:

:$t = \tanh\tfrac12\psi = \frac = \frac$

with the identities

:
\begin
& \sinh\psi = \frac,
& & \cosh\psi = \frac, \\ pt& \tanh\psi = \frac,
& & \coth\psi = \frac, \\ pt& \operatorname\,\psi = \frac,
& & \operatorname\,\psi = \frac,
\end

and

: $e^\psi = \frac, \qquad e^ = \frac.$

Finding in terms of leads to following relationship between the inverse hyperbolic tangent $\operatorname$ and the natural logarithm:

:$2 \operatorname t = \ln\frac.$

The Gudermannian function

Comparing the hyperbolic identities to the circular ones, one notices that they involve the same functions of , just permuted. If we identify the parameter 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

:$\varphi = 2\arctan \bigl(\tanh \tfrac12 \psi\,\bigr) \equiv \operatorname \psi.$

where 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 -axis) give a geometric interpretation of this function.

Rational values and Pythagorean triples

If is a rational number then each of , , , , , and will be a rational number (or be infinite). Likewise if is a rational number then each of , , , , , and will be a rational number (or be infinite). This is helpful with Pythagorean triples; each interior angle has a rational sine because of the SAS area formula for a triangle and has a rational cosine because of the Law of Cosines. These imply that the half-angle tangent is necessarily rational. Vice versa, when a half-angle tangent is a rational number in the interval then the full-angle sine and cosine will both be rational, and there is a right triangle that has the full angle and that has side lengths that are a Pythagorean triple.

Generally, if is a subfield of the complex numbers then implies that . A similar statement can be made about .

See also

*List of trigonometric identities
*Half-side formula

External links

''Tangent Of Halved Angle''
at Planetmath

{{DEFAULTSORT:Tangent Half-Angle Formula
Trigonometry
Conic sections
Mathematical identities