Lösung 4.4:1f - Versionsgeschichte

Tek: Replaced figures with metapost figures

Tek — Tue, 25 Aug 2009 11:25:22 GMT

Replaced figures with metapost figures

← Nächstältere Version		Version vom 11:25, 25. Aug. 2009
Zeile 2:		Zeile 2:

	{\| align="center"		{\| align="center"
-	\| align="center" \|~~[[Image~~:~~4_4_1_f-1~~.~~gif\|center]]~~	+	\| align="center" \|{{:4.4.1f - Solution - The unit circle with angle 2π - π/6}}
	\| width="20px" \|		\| width="20px" \|
-	\| align="center" \|~~[[Image~~:~~4_4_1_f-2~~.~~gif\|center]]~~	+	\| align="center" \|{{:4.4.1f - Solution - The unit circle with angle π + π/6}}
	\|-		\|-
	\| align="center" \|<small>Winkel 2π - π/6 = 11π/6</small>		\| align="center" \|<small>Winkel 2π - π/6 = 11π/6</small>

Bayerer um 14:53, 27. Jul. 2009

Bayerer — Mon, 27 Jul 2009 14:53:55 GMT

← Nächstältere Version		Version vom 14:53, 27. Jul. 2009
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	\| align="center" \|[[Image:4_4_1_f-2.gif\|center]]		\| align="center" \|[[Image:4_4_1_f-2.gif\|center]]
	\|-		\|-
-	\| align="center" \|<small>~~Angle~~ 2π - π/6 = 11π/6</small>	+	\| align="center" \|<small>Winkel 2π - π/6 = 11π/6</small>
	\|\|		\|\|
-	\| align="center" \|<small>~~Angle~~ π + π/6 = 7π/6</small>	+	\| align="center" \|<small>Winkel π + π/6 = 7π/6</small>
	\|}		\|}

	Die beiden Winkel, die <math>\sin v=-1/2</math> erfüllen liegen im dritten und vierten Quadranten, und sind <math>v=2\pi - \pi/6 = 11\pi/6</math> und <math>v = \pi + \pi/6 = 7\pi/6</math>.		Die beiden Winkel, die <math>\sin v=-1/2</math> erfüllen liegen im dritten und vierten Quadranten, und sind <math>v=2\pi - \pi/6 = 11\pi/6</math> und <math>v = \pi + \pi/6 = 7\pi/6</math>.

Markus.bez: Sprache und Formulierung

Markus.bez — Fri, 19 Jun 2009 14:05:38 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 14:05, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir suchen den Winkel der ~~die~~ ''y''-Koordinate <math>-1/2</math> am Einheitskreis entspricht. Vergleichen wir ~~die~~ Übung mit ~~der~~ Übung a, ~~ist~~ es dasselbe Problem, nur dass wir jetzt die Winkel suchen die einen negativen Sinus haben.	+	Wir suchen den Winkel, der der ''y''-Koordinate <math>-1/2</math> am Einheitskreis entspricht. Vergleichen wir diese Übung mit Übung a sehen wir, dass es dasselbe Problem ist, nur dass wir jetzt die Winkel suchen, die einen negativen Sinus haben.

	{\| align="center"		{\| align="center"
Zeile 11:		Zeile 11:
	\|}		\|}

-	Die beiden ~~Winkeln~~ die <math>\sin v=-1/2</math> erfüllen liegen im ~~dridden~~ und vierten Quadranten, und sind <math>v=2\pi - \pi/6 = 11\pi/6</math> und <math>v = \pi + \pi/6 = 7\pi/6</math>.	+	Die beiden Winkel, die <math>\sin v=-1/2</math> erfüllen liegen im dritten und vierten Quadranten, und sind <math>v=2\pi - \pi/6 = 11\pi/6</math> und <math>v = \pi + \pi/6 = 7\pi/6</math>.

Martin um 20:34, 5. Apr. 2009

Martin — Sun, 05 Apr 2009 20:34:58 GMT

← Nächstältere Version		Version vom 20:34, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We can translate the equation <math>\sin v=-1/2</math> to the problem of finding those angles in the unit circle which have a~~ ''y''-~~coordinate of~~ <math>-1/2</math>. ~~If we compare this with the problem that we had in exercise~~ a, ~~where we looked for angles which satisfied <math>\sin v = +1/2</math>, then the situation is the same, except that the angles now lie under, rather than above, the ''x''-axis~~, ~~due to reflectional symmetry~~.	+	Wir suchen den Winkel der die ''y''-Koordinate <math>-1/2</math> am Einheitskreis entspricht. Vergleichen wir die Übung mit der Übung a, ist es dasselbe Problem, nur dass wir jetzt die Winkel suchen die einen negativen Sinus haben.

	{\| align="center"		{\| align="center"
Zeile 11:		Zeile 11:
	\|}		\|}

-	~~The two angles which satisfy~~ <math>\sin v=-1/2</math> ~~lie in the third and fourth quadrants and are~~ <math>v=2\pi - \pi/6 = 11\pi/6</math> ~~and~~ <math>v = \pi + \pi/6 = 7\pi/6</math>.	+	Die beiden Winkeln die <math>\sin v=-1/2</math> erfüllen liegen im dridden und vierten Quadranten, und sind <math>v=2\pi - \pi/6 = 11\pi/6</math> und <math>v = \pi + \pi/6 = 7\pi/6</math>.

Tekbot: hat „Solution 4.4:1f“ nach „Lösung 4.4:1f“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:11:44 GMT

hat „Solution 4.4:1f“ nach „Lösung 4.4:1f“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:11, 22. Okt. 2008

Tek um 13:41, 10. Okt. 2008

Tek — Fri, 10 Oct 2008 13:41:54 GMT

← Nächstältere Version		Version vom 13:41, 10. Okt. 2008
Zeile 1:		Zeile 1:
-	We can translate the equation	+	We can translate the equation <math>\sin v=-1/2</math> to the problem of finding those angles in the unit circle which have a ''y''-coordinate of <math>-1/2</math>. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied <math>\sin v = +1/2</math>, then the situation is the same, except that the angles now lie under, rather than above, the ''x''-axis, due to reflectional symmetry.
-	<math>\sin v=-~~\frac{~~1}{2}</math>	+
-	to the problem of finding those angles in the unit circle which have a y-coordinate of	+
-	<math>-~~\frac{~~1}{2}</math>. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied	+
-	<math>\sin v=+~~\frac{~~1}{2}</math>, then the situation is the same, except that the angles now lie under, rather than above, the	+
-	~~<math>~~x~~</math>~~	+
-	-axis, due to reflectional symmetry.	+

		+	{\| align="center"
		+	\| align="center" \|[[Image:4_4_1_f-1.gif\|center]]
		+	\| width="20px" \|
		+	\| align="center" \|[[Image:4_4_1_f-2.gif\|center]]
		+	\|-
		+	\| align="center" \|<small>Angle 2π - π/6 = 11π/6</small>
		+	\|\|
		+	\| align="center" \|<small>Angle π + π/6 = 7π/6</small>
		+	\|}

-	~~[[Image:4_4_1_f.gif\|center]]~~	+	The two angles which satisfy <math>\sin v=-1/2</math> lie in the third and fourth quadrants and are <math>v=2\pi - \pi/6 = 11\pi/6</math> and <math>v = \pi + \pi/6 = 7\pi/6</math>.
-		+
-	~~Angle~~	+
-	~~<math>2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>~~	+
-	~~Angle~~	+
-	~~<math>\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>~~	+
-		+
-		+
-	The two angles which satisfy	+
-	<math>\sin v=-~~\frac{~~1}{2}</math>	+
-	lie in the third and fourth quadrants and are	+
-	<math>v=2\pi -~~\frac{~~\pi }{6}=~~\frac{~~11\pi }{6}</math>	+
-	and	+
-	<math>v=\pi +~~\frac{~~\pi }{6}=~~\frac{~~7\pi }{6}</math>	+

Ian um 12:47, 30. Sep. 2008

Ian — Tue, 30 Sep 2008 12:47:19 GMT

← Nächstältere Version		Version vom 12:47, 30. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	We can translate the equation
-	<~~center~~> ~~[[Image:4_4_1f.gif]]~~ </~~center~~>	+	<math>\sin v=-\frac{1}{2}</math>
-	{{~~NAVCONTENT_STOP}~~}	+	to the problem of finding those angles in the unit circle which have a y-coordinate of
-		+	<math>-\frac{1}{2}</math>. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied
		+	<math>\sin v=+\frac{1}{2}</math>, then the situation is the same, except that the angles now lie under, rather than above, the
		+	<math>x</math>
		+	-axis, due to reflectional symmetry.


	[[Image:4_4_1_f.gif\|center]]		[[Image:4_4_1_f.gif\|center]]
		+
		+	Angle
		+	<math>2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>
		+	Angle
		+	<math>\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>
		+
		+
		+	The two angles which satisfy
		+	<math>\sin v=-\frac{1}{2}</math>
		+	lie in the third and fourth quadrants and are
		+	<math>v=2\pi -\frac{\pi }{6}=\frac{11\pi }{6}</math>
		+	and
		+	<math>v=\pi +\frac{\pi }{6}=\frac{7\pi }{6}</math>

Tekbot: Lösning 4.4:1f moved to Solution 4.4:1f: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:01:23 GMT

Lösning 4.4:1f moved to Solution 4.4:1f: Robot: moved page

← Nächstältere Version

Version vom 10:01, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:17:49 GMT

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

← Nächstältere Version		Version vom 07:17, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_4_1f.gif]] </center>	+	<center> [[Image:4_4_1f.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}



-	[[~~Bild~~:4_4_1_f.gif\|center]]	+	[[Image:4_4_1_f.gif\|center]]

Madde um 10:35, 19. Aug. 2008

Madde — Tue, 19 Aug 2008 10:35:49 GMT

← Nächstältere Version		Version vom 10:35, 19. Aug. 2008
Zeile 5:		Zeile 5:


-	[[Bild:4_4_1_f.gif\|center~~\| Bildtext??~~]]	+	[[Bild:4_4_1_f.gif\|center]]

← Nächstältere Version		Version vom 11:25, 25. Aug. 2009
Zeile 2:		Zeile 2:

	{\| align="center"		{\| align="center"
-	\| align="center" \|~~[[Image~~:~~4_4_1_f-1~~.~~gif\|center]]~~	+	\| align="center" \|{{:4.4.1f - Solution - The unit circle with angle 2π - π/6}}
	\| width="20px" \|		\| width="20px" \|
-	\| align="center" \|~~[[Image~~:~~4_4_1_f-2~~.~~gif\|center]]~~	+	\| align="center" \|{{:4.4.1f - Solution - The unit circle with angle π + π/6}}
	\|-		\|-
	\| align="center" \|<small>Winkel 2π - π/6 = 11π/6</small>		\| align="center" \|<small>Winkel 2π - π/6 = 11π/6</small>