# derivation of half-angle formulae for tangent

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

Cross-multiply and move terms around:

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

Divide by $\tan(x)$:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Multiply out the numerator and simplify:

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

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

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

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

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

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

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

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

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

Take the square root of both sides to obtain the formula

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

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 DerivationOfHalfangleFormulaeForTangent 2013-03-22 17:00:19 2013-03-22 17:00:19 rspuzio (6075) rspuzio (6075) 9 rspuzio (6075) Derivation msc 26A09 TangentOfHalvedAngle