Derivation of the Half Angle Formulas

Half angle formulas can be derived from the double angle formulas, particularly, the cosine of double angle. For easy reference, the cosines of double angle are listed below:
 

cos 2θ = 1 - 2sin2 θ   →   Equation (1)
cos 2θ = 2cos2 θ - 1   →   Equation (2)

 

Note that the equations above are identities, meaning, the equations are true for any value of the variable θ. The key on the derivation is to substitute θ with ½ θ.
 

For Equations (1) and (2), let θ = ½ θ
 

From Equation (1)
$\cos 2(\frac{1}{2}\theta) = 1 - 2\sin^2 \frac{1}{2}\theta$

$\cos \theta = 1 - 2\sin^2 \frac{1}{2}\theta$

$2\sin^2 \frac{1}{2}\theta = 1 - \cos \theta$

$\sin \frac{1}{2}\theta = \sqrt{\dfrac{1 - \cos \theta}{2}}$

 

From Equation (2)
$\cos 2(\frac{1}{2}\theta) = 2\cos^2 \frac{1}{2}\theta - 1$

$\cos \theta = 2\cos^2 \frac{1}{2}\theta - 1$

$2\cos^2 \frac{1}{2}\theta = 1 + \cos \theta$

$\cos \frac{1}{2}\theta = \sqrt{\dfrac{1 + \cos \theta}{2}}$

 

$\tan \frac{1}{2}\theta = \dfrac{\sin \frac{1}{2}\theta}{\cos \frac{1}{2}\theta}$

$\tan \frac{1}{2}\theta = \dfrac{\sqrt{\dfrac{1 - \cos \theta}{2}}}{\sqrt{\dfrac{1 + \cos \theta}{2}}}$

$\tan \frac{1}{2}\theta = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}$   →   Equation (3)

 

From Equation (3)
$\tan \frac{1}{2}\theta = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta} \times \dfrac{1 - \cos \theta}{1 - \cos \theta}}$

$\tan \frac{1}{2}\theta = \sqrt{\dfrac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}}$

$\tan \frac{1}{2}\theta = \sqrt{\dfrac{(1 - \cos \theta)^2}{\sin^2 \theta}}$

$\tan \frac{1}{2}\theta = \dfrac{1 - \cos \theta}{\sin \theta}$

 

From Equation (3)
$\tan \frac{1}{2}\theta = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta} \times \dfrac{1 + \cos \theta}{1 + \cos \theta}}$

$\tan \frac{1}{2}\theta = \sqrt{\dfrac{1 - \cos^2 \theta}{(1 + \cos \theta)^2}}$

$\tan \frac{1}{2}\theta = \sqrt{\dfrac{\sin^2 \theta}{(1 + \cos \theta)^2}}$

$\tan \frac{1}{2}\theta = \dfrac{\sin \theta}{1 + \cos \theta}$

 

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