Lösung 4.3:3d - Versionsgeschichte

Tek: Replaced figures with metapost figures

Tek — Tue, 25 Aug 2009 07:58:51 GMT

Replaced figures with metapost figures

← Nächstältere Version		Version vom 07:58, 25. Aug. 2009
Zeile 2:		Zeile 2:

	{\| align="center"		{\| align="center"
-	\| align="center" \|~~[[Image~~:~~4_3_3_d-1~~.~~gif]]~~	+	\| align="center" \|{{:4.3.3d - Solution - The unit circle with angle v}}
	\| width="20px"\|		\| width="20px"\|
-	\| align="center" \|~~[[Image~~:~~4_3_3_d~~-2~~.gif]]~~	+	\| align="center" \|{{:4.3.3d - Solution - The unit circle with angle π/2 - v}}
	\|-		\|-
	\| align="center" \|<small>Angle ''v''</small>		\| align="center" \|<small>Angle ''v''</small>

Markus.bez: Sprache und Formulierung

Markus.bez — Fri, 19 Jun 2009 09:25:50 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 09:25, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Der Winkel <math>\pi/2 - v</math> hat denselben Winkel zur positiven ''y''-Achse, wie der Winkel <math>-v</math> zur positiven ''x''-Achse hat.	+	Der Winkel <math>\pi/2 - v</math> hat denselben Winkel zur positiven ''y''-Achse wie der Winkel <math>-v</math> zur positiven ''x''-Achse hat.

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

-	Also hat ~~der Winkel~~ <math>\pi/2 - v</math> ~~dieselbe ''y''-Koordinate,~~ wie die ''x''-Koordinate de Winkels math>v</math>.	+	Also hat die ''y''-Koordinate des Winkels <math>\pi/2 - v</math> denselben Wert wie die ''x''-Koordinate des Winkels <math>v</math>.

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}

-	~~Und von~~ der Übung c wissen wir dass <math>\cos v = \sqrt{1-a^2}\,</math>, also haben wir	+	Von der Übung c wissen wir, dass <math>\cos v = \sqrt{1-a^2}\,</math>, also haben wir

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}

Martin um 12:16, 5. Apr. 2009

Martin — Sun, 05 Apr 2009 12:16:13 GMT

← Nächstältere Version		Version vom 12:16, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The expression for the angle~~ <math>\pi/2 - v</math> ~~differs from <math>\pi/2</math> by as much as <math>-v</math> differs from <math>0</math>. This means that <math>\pi/2-v</math> makes the same angle with the positive~~ ''y''-~~axis as~~ <math>-v</math> ~~makes with the positive~~ ''x''-~~axis~~.	+	Der Winkel <math>\pi/2 - v</math> hat denselben Winkel zur positiven ''y''-Achse, wie der Winkel <math>-v</math> zur positiven ''x''-Achse hat.

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

-	~~Therefore, the angle~~ <math>\pi/2 - v</math> ~~has a~~ ''y''-~~coordinate which is equal to the~~ ''x''-~~coordinate for the angle ''~~v~~'', i.e~~.	+	Also hat der Winkel <math>\pi/2 - v</math> dieselbe ''y''-Koordinate, wie die ''x''-Koordinate de Winkels math>v</math>.

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}

-	~~and from exercise~~ c~~, we know that~~ <math>\cos v = \sqrt{1-a^2}\,</math>,	+	Und von der Übung c wissen wir dass <math>\cos v = \sqrt{1-a^2}\,</math>, also haben wir

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}

Tekbot: hat „Solution 4.3:3d“ nach „Lösung 4.3:3d“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:03:24 GMT

hat „Solution 4.3:3d“ nach „Lösung 4.3:3d“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:03, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:54:28 GMT

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

← Nächstältere Version		Version vom 08:54, 22. Okt. 2008
Zeile 13:		Zeile 13:
	Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.		Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.

-	{{~~Displayed math~~\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}	+	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}

	and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,		and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,

-	{{~~Displayed math~~\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}

Tek um 13:44, 9. Okt. 2008

Tek — Thu, 09 Oct 2008 13:44:01 GMT

← Nächstältere Version		Version vom 13:44, 9. Okt. 2008
Zeile 1:		Zeile 1:
-	The expression for the angle	+	The expression for the angle <math>\pi/2 - v</math> differs from <math>\pi/2</math> by as much as <math>-v</math> differs from <math>0</math>. This means that <math>\pi/2-v</math> makes the same angle with the positive ''y''-axis as <math>-v</math> makes with the positive ''x''-axis.
-	<math>{\pi }/{2~~}\;~~-v</math>	+
-	differs from	+
-	<math>{\pi }/{2~~}\;~~</math>	+
-	by as much as	+
-	<math>-v~~\text{ }~~</math>	+
-	differs from	+
-	<math>0</math>. This means that	+
-	<math>{\pi }/{2~~}\;~~</math>	+
-	makes the same angle with the positive	+
-	~~<math>~~y~~</math>~~	+
-	-axis as	+
-	<math>-v~~\text{ }~~</math>	+
-	makes with the positive	+
-	~~<math>~~x~~</math>~~	+
-	-axis.	+

		+	{\| align="center"
		+	\| align="center" \|[[Image:4_3_3_d-1.gif]]
		+	\| width="20px"\|
		+	\| align="center" \|[[Image:4_3_3_d-2.gif]]
		+	\|-
		+	\| align="center" \|<small>Angle ''v''</small>
		+	\|\|
		+	\| align="center" \|<small>Angle π/2 - ''v''</small>
		+	\|}

-	~~[[Image:4_3_3_d~~.~~gif\|center]]~~	+	Therefore, the angle <math>\pi/2 - v</math> has a ''y''-coordinate which is equal to the ''x''-coordinate for the angle ''v'', i.e.

-	~~Angle~~	+	{{Displayed math\|\|<math>\sin\Bigl(\frac{\pi}{2} - v\Bigr) = \cos v</math>}}
-	<math~~>v</math>~~	+
-	~~angle~~	+
-	<math>\pi -v</math>	+
-		+

-	~~Therefore~~, ~~the angle~~	+	and from exercise c, we know that <math>\cos v = \sqrt{1-a^2}\,</math>,
-	<math>{\~~pi }/~~{2}\~~;-v</math>~~	+
-	~~has a~~	+
-	~~<math>y</math>~~	+
-	~~-coordinate which is equal to the~~	+
-	~~<math>x</math>~~	+
-	~~-coordinate for the angle~~	+
-	~~<math>v~~</math>, ~~i.e.~~	+

-		+	{{Displayed math\|\|<math>\sin\Bigl(\frac{\pi}{2}-v\Bigr) = \sqrt{1-a^2}\,\textrm{.}</math>}}
-	~~<math>\sin \left(~~ {~~\pi }/~~{~~2}\;-v \right)=\cos v</~~math>	+
-		+
-		+
-	~~and from exercise c, we know that~~	+
-	~~<math>\cos v=\sqrt{1-a^{2}}</math>~~	+
-		+
-		+
-		+
-	<math>\sin \~~left~~( \frac{\pi }{2}-v \~~right~~)=\sqrt{1-a^{2}}</math>	+

Ian um 11:11, 29. Sep. 2008

Ian — Mon, 29 Sep 2008 11:11:58 GMT

← Nächstältere Version		Version vom 11:11, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The expression for the angle
-	<~~center~~> ~~[[Image:4_3_3d.gif]]~~ </~~center~~>	+	<math>{\pi }/{2}\;-v</math>
-	{{~~NAVCONTENT_STOP~~}}	+	differs from
		+	<math>{\pi }/{2}\;</math>
		+	by as much as
		+	<math>-v\text{ }</math>
		+	differs from
		+	<math>0</math>. This means that
		+	<math>{\pi }/{2}\;</math>
		+	makes the same angle with the positive
		+	<math>y</math>
		+	-axis as
		+	<math>-v\text{ }</math>
		+	makes with the positive
		+	<math>x</math>
		+	-axis.
		+
		+
	[[Image:4_3_3_d.gif\|center]]		[[Image:4_3_3_d.gif\|center]]
		+
		+	Angle
		+	<math>v</math>
		+	angle
		+	<math>\pi -v</math>
		+
		+
		+	Therefore, the angle
		+	<math>{\pi }/{2}\;-v</math>
		+	has a
		+	<math>y</math>
		+	-coordinate which is equal to the
		+	<math>x</math>
		+	-coordinate for the angle
		+	<math>v</math>, i.e.
		+
		+
		+	<math>\sin \left( {\pi }/{2}\;-v \right)=\cos v</math>
		+
		+
		+	and from exercise c, we know that
		+	<math>\cos v=\sqrt{1-a^{2}}</math>
		+
		+
		+
		+	<math>\sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}</math>

Tekbot: Lösning 4.3:3d moved to Solution 4.3:3d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:53:03 GMT

Lösning 4.3:3d moved to Solution 4.3:3d: Robot: moved page

← Nächstältere Version

Version vom 09:53, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:13:39 GMT

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

← Nächstältere Version		Version vom 07:13, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_3d.gif]] </center>	+	<center> [[Image:4_3_3d.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
-	[[~~Bild~~:4_3_3_d.gif\|center]]	+	[[Image:4_3_3_d.gif\|center]]

Madde um 14:15, 18. Aug. 2008

Madde — Mon, 18 Aug 2008 14:15:58 GMT

← Nächstältere Version		Version vom 14:15, 18. Aug. 2008
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	<center> [[Bild:4_3_3d.gif]] </center>		<center> [[Bild:4_3_3d.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+	[[Bild:4_3_3_d.gif\|center]]

← Nächstältere Version		Version vom 07:58, 25. Aug. 2009
Zeile 2:		Zeile 2:

	{\| align="center"		{\| align="center"
-	\| align="center" \|~~[[Image~~:~~4_3_3_d-1~~.~~gif]]~~	+	\| align="center" \|{{:4.3.3d - Solution - The unit circle with angle v}}
	\| width="20px"\|		\| width="20px"\|
-	\| align="center" \|~~[[Image~~:~~4_3_3_d~~-2~~.gif]]~~	+	\| align="center" \|{{:4.3.3d - Solution - The unit circle with angle π/2 - v}}
	\|-		\|-
	\| align="center" \|<small>Angle ''v''</small>		\| align="center" \|<small>Angle ''v''</small>