derivation of half-angle formulae for tangent

Start with the angle duplication formula

$$\tan (x)=\frac{2\tan (x/2)}{1-{\tan }^2(x/2)}.$$

Cross-multiply and move terms around:

$$\tan (x){\tan }^2(x/2)+2\tan (x/2)=\tan (x)$$

Divide by $\tan (x)$:

$${\tan }^2(x/2)+\frac{2\tan (x/2)}{\tan x}=1$$

Add $1/{\tan }^2(x)$ to both sides:

$${\tan }^2(x/2)+\frac{2\tan (x/2)}{\tan x}+\frac{1}{{\tan }^2(x)}=1+\frac{1}{{\tan }^2(x)}$$

Complete the square (http://planetmath.org/CompletingTheSquare):

$${\left(\tan (x/2)+\frac{1}{\tan (x)}\right)}^2=1+\frac{1}{{\tan }^2(x)}$$

Take a square root and move a term to obtain the half-angle formula:

$$\tan (x/2)=\sqrt{1+\frac{1}{{\tan }^2(x)}}-\frac{1}{\tan (x)}$$

To derive the other forms of the formula, we start by substituting $\sin (x)/\cos (x)$ for $\tan (x)$:

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

Put the stuff inside the square root over a common denominator:

$$\sqrt{\frac{{\sin }^2(x)+{\cos }^2(x)}{{\sin }^2(x)}}-\frac{\cos (x)}{\sin (x)}$$

Recall that ${\sin }^2(x)+{\cos }^2(x)=1$. Hence, we may get rid of the square root:

$$\frac{1}{\sin x}-\frac{\cos (x)}{\sin (x)}$$

Putting the terms over a common denominator, we obtain our formula:

$$\tan (x/2)=\frac{1-\cos (x)}{\sin (x)}$$

To obtain the next formula, multiply both numerator and denominator by $1+\cos (x)$:

$$\frac{(1-\cos (x))(1+\cos (x))}{\sin (x)(1+\cos (x))}$$

Multiply out the numerator and simplify:

$$\frac{1-{\cos }^2(x)}{\sin (x)(1+\cos (x))}$$

Note that the numerator equals ${\sin }^2(x)$:

$$\frac{{\sin }^2(x)}{\sin (x)(1+\cos (x))}$$

Cancel a common factor of $\sin (x)$ to obtain the formula

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

To obtain the last formula, multiply the previous two formulae:

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

Cancel the common factor of $\sin (x)$:

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

Take the square root of both sides to obtain the formula

$$\tan \frac{x}{2}=\pm \sqrt{\frac{1-\cos x}{1+\cos x}};$$

here the sign ($\pm$) has to be chosen according to the quadrant where the angle ${\displaystyle \frac{x}{2}}$ is.

Title	derivation of half-angle formulae for tangent
Canonical name	DerivationOfHalfangleFormulaeForTangent
Date of creation	2013-03-22 17:00:19
Last modified on	2013-03-22 17:00:19
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	9
Author	rspuzio (6075)
Entry type	Derivation
Classification	msc 26A09
Related topic	TangentOfHalvedAngle