# MandysNotes

## 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:

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

## 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:

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

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

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

## Euler’s Formula

13 May 2011

The real part of $e^{i\alpha} \$

is:

$\text{Re}(e^{i \alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2}$

$=\sum_{k=0}^{\infty} (-1)^{k}\frac{\alpha^{2k}}{(2k)!} .$

The imaginary part of $e^{i\alpha} \$

is:

$\text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i}$

$= \sum_{k=1}^{\infty}(-1)^{k -1}\frac{\alpha^{2k -1}}{(2k -1)!} .$

We define:

$\text{Re}(e^{i\alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2} := \cos{\alpha} ;$

$\text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i} := \sin{\alpha}.$

Therefore:

$e^{i\alpha} = \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) =\cos{\alpha} + i \sin{\alpha}.$

This is Euler's Formula.

## Related Rate Problem: Planes Flying

27 January 2010

{module [79]}

Disclaimer: I did not create nor do I own these videos. I have simply embedded them, courtesy of YouTube.

The problem presented in this video is:

A plane is flying at a constant altitude of 2 miles, and at a constant rate of 180mi/hr.  A camera on the ground is following the plane as it flies away from the camera.  How fast must the camera rotate to keep the plane in view when the camera is pointed up at an angle of pi/3?

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