Tangent half-angle formula#Hyperbolic identities

The tangent of half an angle is the stereographic projection of the circle through the point at angle $\backslash pi$ radians onto the line through the angles $\backslash pm\; \backslash frac\{\backslash pi\}\{2\}$. 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 {{mvar|η}} is known to be {{math|0}}, {{math|π/2}}, {{math|π}}, or {{math|3π/2}} because {{math|sin(η)}} and {{math|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}

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 $a=\; \backslash tfrac12\; (\backslash eta+\backslash theta)$ and $b=\; \backslash tfrac12\; (\backslash eta-\backslash theta)$ 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

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

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

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

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

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

File:Tan.half.svg

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

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

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

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

It follows that

$$t\; =\; \backslash frac\{\backslash sin\; \backslash varphi\}\{1+\; \backslash cos\; \backslash varphi\}\; =\; \backslash frac\{\backslash sin\; \backslash varphi(1-\; \backslash cos\; \backslash varphi)\}\{(1+\; \backslash cos\; \backslash varphi)(1-\; \backslash cos\; \backslash varphi)\}\; =\; \backslash frac\{1-\; \backslash cos\; \backslash varphi\}\{\backslash sin\; \backslash varphi\}.$$

{{clear}}

{{Main|Tangent half-angle substitution}}

Image:Weierstrass substitution.svg proof of the tangent half-angle substitution]]

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 {{math|t}} in order to find their antiderivatives.

Geometrically, the construction goes like this: for any point {{math|(cos φ, sin φ)}} on the unit circle, draw the line passing through it and the point {{math|(−1, 0)}}. This point crosses the {{math|y}}-axis at some point {{math|1=y = t}}. One can show using simple geometry that {{math|1=t = tan(φ/2)}}. The equation for the drawn line is {{math|1=y = (1 + x)t}}. The equation for the intersection of the line and circle is then a quadratic equation involving {{math|t}}. The two solutions to this equation are {{math|(−1, 0)}} and {{math|(cos φ, sin φ)}}. This allows us to write the latter as rational functions of {{math|t}} (solutions are given below).

The parameter {{math|t}} represents the stereographic projection of the point {{math|(cos φ, sin φ)}} onto the {{math|y}}-axis with the center of projection at {{math|(−1, 0)}}. Thus, the tangent half-angle formulae give conversions between the stereographic coordinate {{math|t}} on the unit circle and the standard angular coordinate {{math|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

$$e^\{i\; \backslash varphi\}\; =\; \backslash frac\{1\; +\; i\; t\}\{1\; -\; i\; t\},\; \backslash qquad$$

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

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

$$\backslash arctan\; t\; =\; \backslash frac\{-i\}\{2\}\; \backslash ln\backslash frac\{1+it\}\{1-it\}.$$

In calculus, the tangent half-angle substitution is used to find antiderivatives of rational functions of {{math|sin φ}} and {{math|cos φ}}. Differentiating $t=\backslash tan\backslash tfrac12\backslash varphi$ gives

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

and thus

$$d\backslash varphi\; =\; \{\{2\backslash ,dt\}\; \backslash over\; \{1\; +\; t^2\}\}.$$

One can play an entirely analogous game with the hyperbolic functions. A point on (the right branch of) a hyperbola is given by {{math|(cosh ψ, sinh ψ)}}. Projecting this onto {{math|y}}-axis from the center {{math|(−1, 0)}} gives the following:

$$t\; =\; \backslash tanh\backslash tfrac12\backslash psi\; =\; \backslash frac\{\backslash sinh\backslash psi\}\{\backslash cosh\backslash psi+1\}\; =\; \backslash frac\{\backslash cosh\backslash psi-1\}\{\backslash sinh\backslash psi\}$$

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

$$e^\backslash psi\; =\; \backslash frac\{1\; +\; t\}\{1\; -\; t\},\; \backslash qquad$$

e^{-\psi} = \frac{1 - t}{1 + t}.

Finding {{math|ψ}} in terms of {{math|t}} leads to following relationship between the inverse hyperbolic tangent $\backslash operatorname\{artanh\}$ and the natural logarithm:

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

The hyperbolic tangent half-angle substitution in calculus uses

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

{{Main|Gudermannian function}}

Comparing the hyperbolic identities to the circular ones, one notices that they involve the same functions of {{math|t}}, just permuted. If we identify the parameter {{math|t}} in both cases we arrive at a relationship between the circular functions and the hyperbolic ones. That is, if

$$t\; =\; \backslash tan\backslash tfrac12\; \backslash varphi\; =\; \backslash tanh\backslash tfrac12\; \backslash psi$$

then

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

where {{math|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 {{math|y}}-axis) give a geometric interpretation of this function.

{{main article|Pythagorean triple}}

Starting with a Pythagorean triangle with side lengths {{mvar|a}}, {{mvar|b}}, and {{mvar|c}} that are positive integers and satisfy {{math|a{{sup|2}} + b{{sup|2}} {{=}} c{{sup|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 {{math|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 {{mvar|K}} is a subfield of the complex numbers then {{math|tan φ/2 ∈ K ∪ {{mset|∞}}}} implies that {{math|{sin φ, cos φ, tan φ, sec φ, csc φ, cot φ} ⊆ K ∪ {{mset|∞}}}}.

{{Portal|Mathematics}}

• List of trigonometric identities
• Half-side formula

• [http://planetmath.org/encyclopedia/TangentOfHalvedAngle.html Tangent Of Halved Angle] at Planetmath

{{reflist}}

{{DEFAULTSORT:Tangent Half-Angle Formula}}

Category:Trigonometry

Category:Conic sections

Category:Mathematical identities