Tangent half-angle formula

Template:Short description Template:TrigonometryIn 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 $\pi$ radians onto the line through the angles $\pm \frac{\pi }{2}$. Among these formulas are the following:

$$\begin{array}{cc}\tan \frac{1}{2}(\eta \pm \theta )& =\frac{\tan \frac{1}{2}\eta \pm \tan \frac{1}{2}\theta }{1\mp \tan \frac{1}{2}\eta \tan \frac{1}{2}\theta }=\frac{\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }=-\frac{\cos \eta -\cos \theta }{\sin \eta \mp \sin \theta },\\ \tan \frac{1}{2}\theta & =\frac{\sin \theta }{1+\cos \theta }=\frac{\tan \theta }{\sec \theta +1}=\frac{1}{\csc \theta +\cot \theta },& & (\eta =0)\\ \tan \frac{1}{2}\theta & =\frac{1-\cos \theta }{\sin \theta }=\frac{\sec \theta -1}{\tan \theta }=\csc \theta -\cot \theta ,& & (\eta =0)\\ \tan \frac{1}{2}(\theta \pm \frac{1}{2}\pi )& =\frac{1\pm \sin \theta }{\cos \theta }=\sec \theta \pm \tan \theta =\frac{\csc \theta \pm 1}{\cot \theta },& & (\eta =\frac{1}{2}\pi )\\ \tan \frac{1}{2}(\theta \pm \frac{1}{2}\pi )& =\frac{\cos \theta }{1\mp \sin \theta }=\frac{1}{\sec \theta \mp \tan \theta }=\frac{\cot \theta }{\csc \theta \mp 1},& & (\eta =\frac{1}{2}\pi )\\ \frac{1-\tan \frac{1}{2}\theta }{1+\tan \frac{1}{2}\theta }& =\pm \sqrt{\frac{1-\sin \theta }{1+\sin \theta }}\\ \tan \frac{1}{2}\theta & =\pm \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\\ \end{array}$$

Identities

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

$$\begin{array}{cc}\sin \alpha & =\frac{2\tan \frac{1}{2}\alpha }{1+{\tan }^2\frac{1}{2}\alpha }\\ \cos \alpha & =\frac{1-{\tan }^2\frac{1}{2}\alpha }{1+{\tan }^2\frac{1}{2}\alpha }\\ \tan \alpha & =\frac{2\tan \frac{1}{2}\alpha }{1-{\tan }^2\frac{1}{2}\alpha }\end{array}$$

Proofs

Algebraic proofs

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

$$\sin \alpha =2\sin \frac{1}{2}\alpha \cos \frac{1}{2}\alpha =\frac{2\sin \frac{1}{2}\alpha \cos \frac{1}{2}\alpha /{\cos }^2\frac{1}{2}\alpha }{1+{\tan }^2\frac{1}{2}\alpha }=\frac{2\tan \frac{1}{2}\alpha }{1+{\tan }^2\frac{1}{2}\alpha },\text{and}$$

$$\cos \alpha ={\cos }^2\frac{1}{2}\alpha -{\sin }^2\frac{1}{2}\alpha =\frac{({\cos }^2\frac{1}{2}\alpha -{\sin }^2\frac{1}{2}\alpha )/{\cos }^2\frac{1}{2}\alpha }{1+{\tan }^2\frac{1}{2}\alpha }=\frac{1-{\tan }^2\frac{1}{2}\alpha }{1+{\tan }^2\frac{1}{2}\alpha }.$$

Taking the quotient of the formulae for sine and cosine yields

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

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

$$|\sin \alpha |=\sqrt{\frac{1-\cos 2\alpha }{2}}$$ and $$|\cos \alpha |=\sqrt{\frac{1+\cos 2\alpha }{2}}$$

which, upon division gives

$$|\tan \alpha |=\frac{\sqrt{1-\cos 2\alpha }}{\sqrt{1+\cos 2\alpha }}=\frac{\sqrt{1-\cos 2\alpha }\sqrt{1+\cos 2\alpha }}{1+\cos 2\alpha }=\frac{\sqrt{1-{\cos }^{2}2\alpha }}{1+\cos 2\alpha }=\frac{|\sin 2\alpha |}{1+\cos 2\alpha }.$$

Alternatively,

$$|\tan \alpha |=\frac{\sqrt{1-\cos 2\alpha }}{\sqrt{1+\cos 2\alpha }}=\frac{1-\cos 2\alpha }{\sqrt{1+\cos 2\alpha }\sqrt{1-\cos 2\alpha }}=\frac{1-\cos 2\alpha }{\sqrt{1-{\cos }^{2}2\alpha }}=\frac{1-\cos 2\alpha }{|\sin 2\alpha |}.$$

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

$$\begin{array}{cc}\cos (a+b)& =\cos a\cos b-\sin a\sin b\\ \cos (a-b)& =\cos a\cos b+\sin a\sin b\\ \sin (a+b)& =\sin a\cos b+\cos a\sin b\\ \sin (a-b)& =\sin a\cos b-\cos a\sin b\end{array}$$

Pairwise addition of the above four formulae yields:

$$\begin{array}{cc}& \sin (a+b)+\sin (a-b)\\ & =\sin a\cos b+\cos a\sin b+\sin a\cos b-\cos a\sin b\\ & =2\sin a\cos b\\ & \cos (a+b)+\cos (a-b)\\ & =\cos a\cos b-\sin a\sin b+\cos a\cos b+\sin a\sin b\\ & =2\cos a\cos b\end{array}$$

Setting $a=\frac{1}{2}(p+q)$ and $b=\frac{1}{2}(p-q)$ and substituting yields:

$$\begin{array}{cc}& \sin p+\sin q\\ & =\sin (\frac{1}{2}(p+q)+\frac{1}{2}(p-q))+\sin (\frac{1}{2}(p+q)-\frac{1}{2}(p-q))\\ & =2\sin \frac{1}{2}(p+q)\cos \frac{1}{2}(p-q)\\ & \cos p+\cos q\\ & =\cos (\frac{1}{2}(p+q)+\frac{1}{2}(p-q))+\cos (\frac{1}{2}(p+q)-\frac{1}{2}(p-q))\\ & =2\cos \frac{1}{2}(p+q)\cos \frac{1}{2}(p-q)\end{array}$$

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

$$\frac{\sin p+\sin q}{\cos p+\cos q}=\frac{2\sin \frac{1}{2}(p+q)\cos \frac{1}{2}(p-q)}{2\cos \frac{1}{2}(p+q)\cos \frac{1}{2}(p-q)}=\tan \frac{1}{2}(p+q)$$

Geometric proofs

The sides of this rhombus have length 1. The angle between the horizontal line and the shown diagonal is Template:Math. This is a geometric way to prove the particular tangent half-angle formula that says Template:Math. The formulae Template:Math and Template:Math are the ratios of the actual distances to the length of the diagonal.

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

$$\tan \frac{1}{2}(a+b)=\frac{\sin \frac{1}{2}(a+b)}{\cos \frac{1}{2}(a+b)}=\frac{\sin a+\sin b}{\cos a+\cos b}.$$

In the unit circle, application of the above shows that $t=\tan \frac{1}{2}\phi$. By similarity of triangles,

$$\frac{t}{\sin \phi }=\frac{1}{1+\cos \phi }.$$

It follows that

$$t=\frac{\sin \phi }{1+\cos \phi }=\frac{\sin \phi (1-\cos \phi )}{(1+\cos \phi )(1-\cos \phi )}=\frac{1-\cos \phi }{\sin \phi }.$$

The tangent half-angle substitution in integral calculus

A geometric 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 Template:Math in order to find their antiderivatives.

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

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

Then we have

$$\begin{array}{cccc}& \sin \phi =\frac{2t}{1+t^2},& & \cos \phi =\frac{1-t^2}{1+t^2},\\ & \tan \phi =\frac{2t}{1-t^2}& & \cot \phi =\frac{1-t^2}{2t},\\ & \sec \phi =\frac{1+t^2}{1-t^2},& & \csc \phi =\frac{1+t^2}{2t},\end{array}$$

and

$$e^{i\phi }=\frac{1+it}{1-it},e^{-i\phi }=\frac{1-it}{1+it}.$$

Both this expression of $e^{i\phi }$ and the expression $t=\tan (\phi /2)$ can be solved for $\phi$. 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 Template:Math and Template:Math. Differentiating $t=\tan \frac{1}{2}\phi$ gives $$\frac{dt}{d\phi }=\frac{1}{2}{\sec }^2\frac{1}{2}\phi =\frac{1}{2}(1+{\tan }^2\frac{1}{2}\phi )=\frac{1}{2}(1+t^2)$$ and thus $$d\phi =\frac{2dt}{1+t^2}.$$

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

$$t=\tanh \frac{1}{2}\psi =\frac{\sinh \psi }{\cosh \psi +1}=\frac{\cosh \psi -1}{\sinh \psi }$$

with the identities

$$\begin{array}{cccc}& \sinh \psi =\frac{2t}{1-t^2},& & \cosh \psi =\frac{1+t^2}{1-t^2},\\ & \tanh \psi =\frac{2t}{1+t^2},& & \coth \psi =\frac{1+t^2}{2t},\\ & \mathrm{sech}\psi =\frac{1-t^2}{1+t^2},& & \mathrm{csch}\psi =\frac{1-t^2}{2t},\end{array}$$

and

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

Finding Template:Math in terms of Template:Math leads to following relationship between the inverse hyperbolic tangent $\mathrm{artanh}$ and the natural logarithm:

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

The hyperbolic tangent half-angle substitution in calculus uses $$d\psi =\frac{2dt}{1-t^2}.$$

The Gudermannian function

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

$$t=\tan \frac{1}{2}\phi =\tanh \frac{1}{2}\psi$$

then

$$\phi =2\arctan (\tanh \frac{1}{2}\psi )\equiv \mathrm{gd}\psi .$$

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

Rational values and Pythagorean triples

Template:Main article Starting with a Pythagorean triangle with side lengths Template:Mvar, Template:Mvar, and Template:Mvar that are positive integers and satisfy Template:Math, 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 Template:Math.

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 Template:Mvar is a subfield of the complex numbers then Template:Math implies that Template:Math.

See also

• List of trigonometric identities
• Half-side formula

External links

• Tangent Of Halved Angle at Planetmath