<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>23.6.2.3 Revision of the exponential form of a complex number</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[Adding the results from the previous task gives
\[e^{i\pi/6} + e^{-i\pi/6} = 2\cos \left( \dfrac{\pi}{6} \right) \] 
or
\[ \cos \left( \dfrac{\pi}{6} \right) = \dfrac{1}{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right) .\]
Subtracting the results from the previous task gives
\[e^{i\pi/6} - e^{-i\pi/6} = 2 i \sin \left( \dfrac{\pi}{6} \right) \] 
or
\[ \sin \left( \dfrac{\pi}{6} \right) = \dfrac{1}{2i} \left( e^{i\pi/6} - e^{-i\pi/6} \right) .\]
Don't forget the factor \(i\) in this latter case.
<p></p>
Clearly, similar calculations could be carried out for any angle \(\theta\). The general results are summarised
in the following Key Point.
<div class="HELM_keypoint">
<h4>Key Point</h4>
<p><b>Euler's Relations</b></p>
\[ e^{i\theta} \equiv \cos\theta  + i \sin \theta, \hspace{1cm} e^{-i\theta} \equiv \cos\theta  - i \sin \theta\]
\[\cos \theta \equiv \dfrac{1}{2} ( e^{i\theta} + e^{-i\theta} ), \hspace{1cm} \sin \theta \equiv \dfrac{1}{2i} ( e^{i\theta} - e^{-i\theta} )\]
</div>
<p></p>
Using these results we can redraft an expression of the form
\[ a_n \cos n\theta + b_n \sin n\theta \]
in terms of complex exponentials. 
<p></p>
(This expression, with \(\omega = \omega_0 t\), is of course the \(n^{ \text{th} }\) harmonic of a trigonometric Fourier series.)]]></text>
    </questiontext>
    <generalfeedback format="html">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
