# MandysNotes

## Half Angle Formulae

21 May 2011

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:

$\gamma = 2\alpha ,$

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

## Addition and Double Angle Formulae

20 May 2011

From the definition of cosine:

$\cos{\theta} =\frac{e^{i\theta} + e^{-i\theta}}{2}$

we find:

$2\cos{(\alpha + \beta)} = e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} .$

## The Pythagorean Theorem from Euler’s Formula

14 May 2011

Consider any complex number $z, \$with modulus

$\sqrt{z\bar{z} } = |z| = r \ ,$

where $r \in \mathbb{R^{+}}\ .$

Note that $\frac{z}{r} \$would be a complex number with modulus one:

$| \frac{z}{r} | = | \frac{z}{|z|} | = 1.$

## Sum of Squares Using Euler's Formula

14 May 2011

Euler's formula makes life much easier when it comes to deriving trigonometric identities.

Note, first of all, that $|e^{i\alpha}| = 1,$gives us:

$\left( \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) \right) \left( \text{Re}(e^{i\alpha}) – i\text{Im}(e^{i\alpha}) \right)$

$= \left( \cos{\alpha} + i\sin{\alpha} \right) \left( \cos{\alpha} - i\sin{\alpha} \right)$

$= \cos^{2}{\alpha} + \sin^{2}{\alpha} = 1.$

Page 1 of 2