Lösung 4.3:3f - Versionsgeschichte

Markus.bez um 09:44, 19. Jun. 2009

2009-06-19T09:44:23Z

← Nächstältere Version		Version vom 09:44, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Hier ist es schwierig Geometrie zu verwenden~~, und wir~~ betrachten stattdessen das Additionstheorem	+	Hier ist es schwierig Geometrie zu verwenden. Wir betrachten stattdessen das Additionstheorem

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}

-	Nachdem <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, und <math>\cos v=\sqrt{1-a^2}</math> erhalten wir	+	Nachdem <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, und <math>\cos v=\sqrt{1-a^2}</math>, erhalten wir

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}

2009-04-05T12:22:15Z

← Nächstältere Version		Version vom 12:22, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~In this case, it is perhaps simplest to use the addition formula for sine~~,	+	Hier ist es schwierig Geometrie zu verwenden, und wir betrachten stattdessen das Additionstheorem

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}

-	~~Since~~ <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, ~~and~~ <math>\cos v=\sqrt{1-a^2}</math> ~~this can be written as~~	+	Nachdem <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, und <math>\cos v=\sqrt{1-a^2}</math> erhalten wir

	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}

2008-10-22T15:04:04Z

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

← Nächstältere Version

Version vom 15:04, 22. Okt. 2008

2008-10-22T08:54:48Z

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

← Nächstältere Version		Version vom 08:54, 22. Okt. 2008
Zeile 1:		Zeile 1:
	In this case, it is perhaps simplest to use the addition formula for sine,		In this case, it is perhaps simplest to use the addition formula for sine,

-	{{~~Displayed math~~\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}

	Since <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, and <math>\cos v=\sqrt{1-a^2}</math> this can be written as		Since <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, and <math>\cos v=\sqrt{1-a^2}</math> this can be written as

-	{{~~Displayed math~~\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}

2008-10-09T14:03:35Z

← Nächstältere Version		Version vom 14:03, 9. Okt. 2008
Zeile 1:		Zeile 1:
	In this case, it is perhaps simplest to use the addition formula for sine,		In this case, it is perhaps simplest to use the addition formula for sine,

		+	{{Displayed math\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \sin\frac{\pi }{3}\cdot \cos v + \cos\frac{\pi}{3}\cdot\sin v\,\textrm{.}</math>}}

-	<math>\sin ~~\left~~( ~~\frac{~~\pi }{3~~}+v \right~~)=\~~sin \frac{\pi }~~{3}\~~centerdot~~ \cos ~~v+\cos \frac{~~\pi }{3~~}\centerdot~~ \sin v.</math>	+	Since <math>\sin (\pi/3) = \sqrt{3}/\!2</math>, <math>\cos (\pi/3) = 1/2</math>, <math>\sin v = a</math>, and <math>\cos v=\sqrt{1-a^2}</math> this can be written as

-	~~Since~~	+	{{Displayed math\|\|<math>\sin\Bigl(\frac{\pi}{3}+v\Bigr) = \frac{\sqrt{3}}{2}\sqrt{1-a^2} + \frac{1}{2}a\,\textrm{.}</math>}}
-	~~<math>\sin \frac~~{~~\pi }~~{~~3}=\frac{\sqrt{3}}{2},\ \ \cos \frac{\pi }{3}=\frac{1}{2},\ \ \sin v=a</~~math~~>, and~~	+
-	~~<math>\cos v=\sqrt{1-a^{2}}</math>~~	+
-	~~this can be written as~~	+
-		+
-		+
-	<math>\sin \~~left~~( \frac{\pi }{3}+v \~~right~~)=\frac{\sqrt{3}}{2}\sqrt{1-a^{2}}+\frac{1}{2}a.</math>	+

2008-09-29T11:34:42Z

← Nächstältere Version		Version vom 11:34, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	In this case, it is perhaps simplest to use the addition formula for sine,
-	<~~center~~> ~~[[Image:4_3_3f.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\sin \left( \frac{\pi }{3}+v \right)=\sin \frac{\pi }{3}\centerdot \cos v+\cos \frac{\pi }{3}\centerdot \sin v.</math>
		+
		+	Since
		+	<math>\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2},\ \ \cos \frac{\pi }{3}=\frac{1}{2},\ \ \sin v=a</math>, and
		+	<math>\cos v=\sqrt{1-a^{2}}</math>
		+	this can be written as
		+
		+
		+	<math>\sin \left( \frac{\pi }{3}+v \right)=\frac{\sqrt{3}}{2}\sqrt{1-a^{2}}+\frac{1}{2}a.</math>

2008-09-09T09:53:43Z

← Nächstältere Version

Version vom 09:53, 9. Sep. 2008

2008-08-21T07:13:59Z

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

2008-04-03T09:46:24Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_3f.gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:4_3_3f.gif]] </center>
{{NAVCONTENT_STOP}}