# MandysNotes

Saturday, 21 May 2011 00:00

## Half Angle Formulae

By  Gideon
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We can find the half angle formula for cosine by starting with the double angle formula:

$\cos{(2\alpha)} = \cos^{2}{\alpha} - \sin^{2}{\alpha} \ .$

Making the substitution:

$\gamma = 2\alpha ,$

$\cos{\gamma} = \cos^{2}{(\frac{\gamma}{2})} - \sin^{2}{(\frac{\gamma}{2})} .$

Now we use the relation:

$\cos^{2}{(\theta)} + \sin^{2}{(\theta)} = 1 ,$

$\cos^{2}{(\theta)} - 1 = - \sin^{2}{(\theta)} .$

(valid for any $$\theta \$$)

to obtain:

$\cos{\gamma} = \cos^{2}{(\frac{\gamma}{2})} + \cos^{2}{(\frac{\gamma}{2})} - 1$

$= 2\cos^{2}{\frac{\gamma}{2}} \ -1 .$

Solving this for $\cos{(\frac{\gamma}{2})} \ ,$

gives:

$\cos{(\frac{\gamma}{2})} = \pm \sqrt{\frac{ \cos{\gamma} + 1}{2}} .$

Where the plus sign is appropriate for $0 \leq \gamma \leq \pi \ ,$

and the minus sign for $\pi \leq \gamma \leq 2\pi \ .$

To find the half angle formula for sine, we use the fact that:

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

to obtain:

$\sin{(\frac{\gamma}{2})}$

$= \pm \sqrt{ 1 - \frac{\cos{\gamma}}{2} - \frac{1}{2} }$

$= \pm \sqrt{\frac{ 1 - \cos{\gamma}}{2}} .$