Lösung 4.2:4f - Versionsgeschichte

Markus.bez um 13:09, 18. Jun. 2009

Markus.bez — Thu, 18 Jun 2009 13:09:24 GMT

← Nächstältere Version		Version vom 13:09, 18. Jun. 2009
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-	Wir addieren <math>2\pi</math> zu <math>-5\pi/3\,</math>, sodass wir einen Winkel im ersten Quadrant erhalten der denselben Tangens hat wie <math>-5\pi/3</math>,	+	Wir addieren <math>2\pi</math> zu <math>-5\pi/3\,</math>, sodass wir einen Winkel im ersten Quadrant erhalten, der denselben Tangens hat wie <math>-5\pi/3</math>:

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Martin — Sat, 04 Apr 2009 15:54:19 GMT

← Nächstältere Version		Version vom 15:54, 4. Apr. 2009
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-	~~If we add~~ <math>2\pi</math> to <math>-5\pi/3\,</math>, ~~we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle~~ <math>-5\pi/3</math> ~~and consequently has the same tangent value~~,	+	Wir addieren <math>2\pi</math> zu <math>-5\pi/3\,</math>, sodass wir einen Winkel im ersten Quadrant erhalten der denselben Tangens hat wie <math>-5\pi/3</math>,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot — Wed, 22 Oct 2008 14:57:44 GMT

hat „Solution 4.2:4f“ nach „Lösung 4.2:4f“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:57, 22. Okt. 2008

Tekbot — Wed, 22 Oct 2008 08:52:48 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:52, 22. Okt. 2008
Zeile 1:		Zeile 1:
	If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value,		If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\tan\Bigl(-\frac{5\pi}{3}\Bigr)		\tan\Bigl(-\frac{5\pi}{3}\Bigr)
	= \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr)		= \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr)

Tek — Thu, 09 Oct 2008 10:52:31 GMT

← Nächstältere Version		Version vom 10:52, 9. Okt. 2008
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-	If we add	+	If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value,
-	<math>2\pi </math>	+
-	to	+
-	<math>-~~\frac{~~5\pi }{3}</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle	+
-	<math>-~~\frac{~~5\pi }{3}</math>	+
-	and consequently has the same tangent value:	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\tan\Bigl(-\frac{5\pi}{3}\Bigr)
-	& \tan \~~left~~( -\frac{5\pi }{3} \~~right~~)=\tan \~~left~~( -\frac{5\pi }{3}+2\pi \~~right~~)=\tan \frac{\pi }{3} \\	+	= \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr)
-	& =\frac{\sin \~~frac~~{\pi }{3}}{\cos \~~frac~~{\pi }{3}}=\frac{\~~frac~~{\sqrt{3}}{2}}{\~~frac~~{1}{2}}=\sqrt{3} \\	+	= \tan\frac{\pi}{3}
-	\end{align}</math>	+	= \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}}
		+	= \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}
		+	= \sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}

Ian — Sun, 28 Sep 2008 13:27:46 GMT

← Nächstältere Version		Version vom 13:27, 28. Sep. 2008
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-	{{~~NAVCONTENT_START}~~}	+	If we add
-	<~~center~~> ~~[[Image:4_2_4f.gif]]~~ </~~center~~>	+	<math>2\pi </math>
-	{{~~NAVCONTENT_STOP~~}}	+	to
		+	<math>-\frac{5\pi }{3}</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle
		+	<math>-\frac{5\pi }{3}</math>
		+	and consequently has the same tangent value:
		+
		+
		+	<math>\begin{align}
		+	& \tan \left( -\frac{5\pi }{3} \right)=\tan \left( -\frac{5\pi }{3}+2\pi \right)=\tan \frac{\pi }{3} \\
		+	& =\frac{\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \\
		+	\end{align}</math>

Tekbot — Tue, 09 Sep 2008 09:47:23 GMT

← Nächstältere Version

Version vom 09:47, 9. Sep. 2008

Tekbot — Thu, 21 Aug 2008 07:10:49 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

Oskar — Thu, 03 Apr 2008 08:02:42 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_4f.gif </center> {{NAVCONTENT_STOP}}

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