<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:wfw="http://wellformedweb.org/CommentAPI/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:slash="http://purl.org/rss/1.0/modules/slash/"
	>

<channel>
	<title>Dianemarie &#8211; Die Kluge Eule</title>
	<atom:link href="https://dieklugeeule.com/author/dianemarie/feed/" rel="self" type="application/rss+xml" />
	<link>https://dieklugeeule.com</link>
	<description></description>
	<lastBuildDate>Wed, 12 Feb 2020 18:46:36 +0000</lastBuildDate>
	<language>de-DE</language>
	<sy:updatePeriod>
	hourly	</sy:updatePeriod>
	<sy:updateFrequency>
	1	</sy:updateFrequency>
	<generator>https://wordpress.org/?v=6.0.1</generator>

<image>
	<url>https://dieklugeeule.com/wp-content/uploads/2022/04/cropped-logo-smal-2-32x32.jpg</url>
	<title>Dianemarie &#8211; Die Kluge Eule</title>
	<link>https://dieklugeeule.com</link>
	<width>32</width>
	<height>32</height>
</image> 
	<item>
		<title>Wie löst man die Identität #cos3x = 4cos ^ 3x &#8211; 3cosx #?</title>
		<link>https://dieklugeeule.com/wie-lost-man-die-identitat-cos3x-4cos-3x-3cosx/</link>
		
		<dc:creator><![CDATA[Dianemarie]]></dc:creator>
		<pubDate>Wed, 12 Feb 2020 18:46:36 +0000</pubDate>
				<category><![CDATA[Trigonometrie]]></category>
		<guid isPermaLink="false">https://dieklugeeule.com/?p=9748</guid>

					<description><![CDATA[Wie löst man die Identität #cos3x = 4cos ^ 3x - 3cosx #? Antworten: Siehe Erklärung. Erläuterung: Also werden wir das beweisen #cos3x=4cos^3x-3cosx# #[1]color(white)(XX)cos3x# #[2]color(white)(XX)=cos(x+2x)# Winkelsummenidentität: #cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta# #[3]color(white)(XX)=cosxcos2x-sinxsin2x# Doppelte Winkelidentität: #cos2alpha=2cos^2alpha-1# #[4]color(white)(XX)=cosx(2cos^2x-1)-sinxsin2x# #[5]color(white)(XX)=2cos^3x-cosx-sinxsin2x# Doppelte Winkelidentität: #sin2alpha=2sinalphacosalpha# #[6]color(white)(XX)=2cos^3x-cosx-sinx(2sinxcosx)# #[7]color(white)(XX)=2cos^3x-cosx-sin^2x(2cosx)# Pythagoreische Identität: #sin^2alpha=1-cos^2alpha# #[8]color(white)(XX)=2cos^3x-cosx-(1-cos^2x)(2cosx)# #[9]color(white)(XX)=2cos^3x-cosx-(2cosx-2cos^3x)# #[10]color(white)(XX)=2cos^3x-cosx-2cosx+2cos^3x# Kombiniere gleiche Begriffe. #[11]color(white)(XX)=4cos^3x-3cosx# #color(blue)( :.cos3x=4cos^3x-3cosx)#]]></description>
										<content:encoded><![CDATA[<h1 class="questionTitle">Wie löst man die Identität #cos3x = 4cos ^ 3x - 3cosx #?</h1>
<div class="answerContainer clearfix">
<div class='answerText'>
<div class="answerSummary">
<h4 class="answerHeader">Antworten:</h4>
<div>
<div class='markdown'>
<p>Siehe Erklärung.</p>
</div></div>
</p></div>
<div class="answerDescription">
<h4 class="answerHeader">Erläuterung:</h4>
<div>
<div class='markdown'>
<p class="gt-block"><strong>Also werden wir das beweisen #cos3x=4cos^3x-3cosx#</strong></p>
<p>#[1]color(white)(XX)cos3x#</p>
<p>#[2]color(white)(XX)=cos(x+2x)#</p>
<p class="gt-block"><em>Winkelsummenidentität: #cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta#</em></p>
<p>#[3]color(white)(XX)=cosxcos2x-sinxsin2x#</p>
<p class="gt-block"><em>Doppelte Winkelidentität: #cos2alpha=2cos^2alpha-1#</em></p>
<p>#[4]color(white)(XX)=cosx(2cos^2x-1)-sinxsin2x#</p>
<p>#[5]color(white)(XX)=2cos^3x-cosx-sinxsin2x#</p>
<p class="gt-block"><em>Doppelte Winkelidentität: #sin2alpha=2sinalphacosalpha#</em></p>
<p>#[6]color(white)(XX)=2cos^3x-cosx-sinx(2sinxcosx)#</p>
<p>#[7]color(white)(XX)=2cos^3x-cosx-sin^2x(2cosx)#</p>
<p class="gt-block"><em>Pythagoreische Identität: #sin^2alpha=1-cos^2alpha#</em></p>
<p>#[8]color(white)(XX)=2cos^3x-cosx-(1-cos^2x)(2cosx)#</p>
<p>#[9]color(white)(XX)=2cos^3x-cosx-(2cosx-2cos^3x)#</p>
<p>#[10]color(white)(XX)=2cos^3x-cosx-2cosx+2cos^3x#</p>
<p class="gt-block"><em>Kombiniere gleiche Begriffe.</em></p>
<p>#[11]color(white)(XX)=4cos^3x-3cosx#</p>
<p>#color(blue)( :.cos3x=4cos^3x-3cosx)#</p>
</div></div>
</p></div>
</p></div>
</p></div>
]]></content:encoded>
					
		
		
			</item>
	</channel>
</rss>
