Lösung 4.2:4a - Versionsgeschichte

Tek: Replaced figure with metapost figure

2009-08-24T12:27:09Z

Replaced figure with metapost figure

← Nächstältere Version		Version vom 12:27, 24. Aug. 2009
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

-	~~[[Image~~:~~4_2_4_a~~.~~gif\|~~center]]	+	<center>{{:4.2.4a - Solution - Two unit circles with angles 11π/6 and -π/6, respectively}}</center>

Markus.bez um 13:05, 18. Jun. 2009

2009-06-18T13:05:57Z

← Nächstältere Version		Version vom 13:05, 18. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir schreiben <math>11\pi/6</math> ~~wie~~	+	Wir schreiben <math>11\pi/6</math> als

	{{Abgesetzte Formel\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}

-	und sehen dass der Winkel im vierten Quadrant liegt.	+	und sehen, dass der Winkel im vierten Quadrant liegt.

-	Wir sehen auch dass der ~~winkel denselben~~ Punkt am Einheitskreis entspricht, wie der Winkel <math>\cos (-\pi/6)</math>, den wir schon in der Übung 4.2:3f berechnet haben. Also haben wir	+	Wir sehen auch, dass der Winkel demselben Punkt am Einheitskreis entspricht wie der Winkel <math>\cos (-\pi/6)</math>, den wir schon in der Übung 4.2:3f berechnet haben. Also haben wir

	{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

Martin um 14:41, 4. Apr. 2009

2009-04-04T14:41:08Z

← Nächstältere Version		Version vom 14:41, 4. Apr. 2009
Zeile 1:		Zeile 1:
-	~~It can be a little difficult to draw the angle~~ <math>11\pi/6</math> ~~straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as~~	+	Wir schreiben <math>11\pi/6</math> wie

	{{Abgesetzte Formel\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}

-	~~we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left~~.	+	und sehen dass der Winkel im vierten Quadrant liegt.

-	~~We also note that this angle corresponds to exactly the same point on the unit circle as the angle <math>-\pi/6</math>~~, ~~and because we calculated~~ <math>\cos (-\pi/6)</math> in ~~exercise~~ 4.2:3f~~, we have that~~	+	Wir sehen auch dass der winkel denselben Punkt am Einheitskreis entspricht, wie der Winkel <math>\cos (-\pi/6)</math>, den wir schon in der Übung 4.2:3f berechnet haben. Also haben wir

	{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

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

2008-10-22T14:56:04Z

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

← Nächstältere Version

Version vom 14:56, 22. Okt. 2008

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

2008-10-22T08:51:58Z

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

← Nächstältere Version		Version vom 08:51, 22. Okt. 2008
Zeile 1:		Zeile 1:
	It can be a little difficult to draw the angle <math>11\pi/6</math> straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as		It can be a little difficult to draw the angle <math>11\pi/6</math> straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as

-	{{~~Displayed math~~\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}

	we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.		we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.
Zeile 7:		Zeile 7:
	We also note that this angle corresponds to exactly the same point on the unit circle as the angle <math>-\pi/6</math>, and because we calculated <math>\cos (-\pi/6)</math> in exercise 4.2:3f, we have that		We also note that this angle corresponds to exactly the same point on the unit circle as the angle <math>-\pi/6</math>, and because we calculated <math>\cos (-\pi/6)</math> in exercise 4.2:3f, we have that

-	{{~~Displayed math~~\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

Tek um 08:27, 9. Okt. 2008

2008-10-09T08:27:04Z

← Nächstältere Version		Version vom 08:27, 9. Okt. 2008
Zeile 1:		Zeile 1:
-	It can be a little difficult to draw the angle	+	It can be a little difficult to draw the angle <math>11\pi/6</math> straight onto a unit circle, but if we rewrite <math>11\pi/6</math> as
-	<math>~~\frac{~~11\pi }{6}</math>	+
-	straight onto a unit circle, but if we rewrite	+
-	<math>~~\frac{~~11\pi }{6}</math>	+
-		+
-	as	+
-		+
-	~~<math>\frac{11\pi }{6}=\frac{6\pi +3\pi +2\pi }{6}=\pi +\frac{\pi }{2}+\frac{\pi }{3}</math>~~	+

		+	{{Displayed math\|\|<math>\frac{11\pi}{6} = \frac{6\pi+3\pi+2\pi}{6} = \pi + \frac{\pi}{2} + \frac{\pi}{3}</math>}}

	we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.		we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.

-	We also note that this angle corresponds to exactly the same point on the unit circle as the angle	+	We also note that this angle corresponds to exactly the same point on the unit circle as the angle <math>-\pi/6</math>, and because we calculated <math>\cos (-\pi/6)</math> in exercise 4.2:3f, we have that
-	<math>-~~\frac{~~\pi }{6}</math>, and because we calculated	+
-	<math>\cos ~~\left~~( -~~\frac{~~\pi }{6~~} \right~~)</math>	+
-	in exercise f, we have that	+
-		+
-		+
-	~~<math>\cos \frac{11\pi }{6}=\cos \left( -\frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}</math>~~	+

		+	{{Displayed math\|\|<math>\cos\frac{11\pi}{6} = \cos\Bigl(-\frac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}\,\textrm{.}</math>}}

	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

Ian um 12:52, 28. Sep. 2008

2008-09-28T12:52:21Z

← Nächstältere Version		Version vom 12:52, 28. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	It can be a little difficult to draw the angle
-	<~~center~~> ~~[[Image:4_2_4a~~.~~gif]]~~ </~~center~~>	+	<math>\frac{11\pi }{6}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	straight onto a unit circle, but if we rewrite
		+	<math>\frac{11\pi }{6}</math>
		+
		+	as
		+
		+	<math>\frac{11\pi }{6}=\frac{6\pi +3\pi +2\pi }{6}=\pi +\frac{\pi }{2}+\frac{\pi }{3}</math>
		+
		+
		+	we see that we have an angle that lies in the fourth quadrant, as in the figure below to the left.
		+
		+	We also note that this angle corresponds to exactly the same point on the unit circle as the angle
		+	<math>-\frac{\pi }{6}</math>, and because we calculated
		+	<math>\cos \left( -\frac{\pi }{6} \right)</math>
		+	in exercise f, we have that
		+
		+
		+	<math>\cos \frac{11\pi }{6}=\cos \left( -\frac{\pi }{6} \right)=\frac{\sqrt{3}}{2}</math>
		+
		+
	[[Image:4_2_4_a.gif\|center]]		[[Image:4_2_4_a.gif\|center]]

Tekbot: Lösning 4.2:4a moved to Solution 4.2:4a: Robot: moved page

2008-09-09T09:45:43Z

Lösning 4.2:4a moved to Solution 4.2:4a: Robot: moved page

← Nächstältere Version

Version vom 09:45, 9. Sep. 2008

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

2008-08-21T07:09:59Z

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

← Nächstältere Version		Version vom 07:09, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_2_4a.gif]] </center>	+	<center> [[Image:4_2_4a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[~~Bild~~:4_2_4_a.gif\|center]]	+	[[Image:4_2_4_a.gif\|center]]

Madde um 15:05, 15. Aug. 2008

2008-08-15T15:05:55Z

← Nächstältere Version		Version vom 15:05, 15. Aug. 2008
Zeile 2:		Zeile 2:
	<center> [[Bild:4_2_4a.gif]] </center>		<center> [[Bild:4_2_4a.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	[[Bild:4_2_4_a.gif\|center]]