MandysNotes

Saturday, 21 May 2011 00:00

Half Angle Formulae

By  Gideon
Rate this item
(0 votes)

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:

\[ \gamma = 2\alpha ,\]

leads to the equation:

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

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:

\[ \sin{(\frac{\gamma}{2})} \]

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

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

Read 3093 times Last modified on Tuesday, 01 April 2014 19:14
Login to post comments

NOTRad