Lösung 4.3:8c - Versionsgeschichte

Bayerer um 10:48, 24. Jul. 2009

2009-07-24T10:48:52Z

← Nächstältere Version		Version vom 10:48, 24. Jul. 2009
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}

-	~~Den~~ 1~~-er~~ im Nenner können wir ~~wie~~ <math>\cos^2(u/2) + \sin^2(u/2)</math> schreiben, und so erhalten wir	+	Die 1 im Nenner können wir als <math>\cos^2(u/2) + \sin^2(u/2)</math> schreiben, und so erhalten wir

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

Markus.bez um 13:09, 19. Jun. 2009

2009-06-19T13:09:35Z

← Nächstältere Version		Version vom 13:09, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir könnten <math>\tan\frac{u}{2}</math> ~~wie ein~~ Bruch mit Sinus und Kosinus schreiben, und ~~also~~ den Ausdruck	+	Wir könnten <math>\tan\frac{u}{2}</math> als Bruch mit Sinus und Kosinus schreiben und dann den Ausdruck

	{{Abgesetzte Formel\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}		{{Abgesetzte Formel\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}

-	erweitern. Dies ergibt aber Wurzelausdrücke, die recht komplizierte Rechnungen ~~ergeben~~. Stattdessen ~~betrachden~~ wir die rechte Seite der Gleichung, und erweitern diese.	+	erweitern. Dies ergibt aber Wurzelausdrücke, die recht komplizierte Rechnungen erfordern. Stattdessen betrachten wir die rechte Seite der Gleichung und erweitern diese.

	Wir schreiben <math>u</math> wie <math>2\cdot(u/2)</math> und verwenden die Doppelwinkelfunktion,		Wir schreiben <math>u</math> wie <math>2\cdot(u/2)</math> und verwenden die Doppelwinkelfunktion,
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}

-	Den 1:er im Nenner können wir wie <math>\cos^2(u/2) + \sin^2(u/2)</math> schreiben, und so erhalten wir	+	Den 1-er im Nenner können wir wie <math>\cos^2(u/2) + \sin^2(u/2)</math> schreiben, und so erhalten wir

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

Martin um 13:21, 5. Apr. 2009

2009-04-05T13:21:21Z

← Nächstältere Version		Version vom 13:21, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~One could write~~ <math>\tan\frac{u}{2}</math> ~~as a quotient involving sine and cosine, and then continue with the formula for half-angles~~,	+	Wir könnten <math>\tan\frac{u}{2}</math> wie ein Bruch mit Sinus und Kosinus schreiben, und also den Ausdruck

	{{Abgesetzte Formel\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}		{{Abgesetzte Formel\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}

-	~~but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots~~, ~~it is perhaps simpler instead to go backwards and work with the right-hand side~~.	+	erweitern. Dies ergibt aber Wurzelausdrücke, die recht komplizierte Rechnungen ergeben. Stattdessen betrachden wir die rechte Seite der Gleichung, und erweitern diese.

-	~~We write~~ <math>u</math> as <math>2\cdot(u/2)</math>~~and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument)~~,	+	Wir schreiben <math>u</math> wie <math>2\cdot(u/2)</math> und verwenden die Doppelwinkelfunktion,

	{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}

-	~~Writing the~~ 1 ~~in the denominator as~~ <math>\cos^2(u/2) + \sin^2(u/2)</math> ~~using the Pythagorean identity~~,	+	Den 1:er im Nenner können wir wie <math>\cos^2(u/2) + \sin^2(u/2)</math> schreiben, und so erhalten wir

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

Tekbot: hat „Solution 4.3:8c“ nach „Lösung 4.3:8c“ verschoben: Robot: moved page

2008-10-22T15:09:04Z

hat „Solution 4.3:8c“ nach „Lösung 4.3:8c“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:09, 22. Okt. 2008

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

2008-10-22T08:57:18Z

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

← Nächstältere Version		Version vom 08:57, 22. Okt. 2008
Zeile 1:		Zeile 1:
	One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles,		One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles,

-	{{~~Displayed math~~\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}	+	{{Abgesetzte Formel\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}

	but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.		but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.
Zeile 7:		Zeile 7:
	We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument),		We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument),

-	{{~~Displayed math~~\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}

	Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,		Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}		\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}
	&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt]		&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt]

Tek um 08:38, 10. Okt. 2008

2008-10-10T08:38:31Z

← Nächstältere Version		Version vom 08:38, 10. Okt. 2008
Zeile 1:		Zeile 1:
-	One could write	+	One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles,
-	<math>\tan \frac{u}{2}</math>	+
-	as a quotient involving sine and cosine, and then continue with the formula for half-angles,	+
-		+
-		+
-	~~<math>\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...</math>~~	+

		+	{{Displayed math\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}

	but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.		but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.

-	We write	+	We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument),
-	<math>u</math>	+
-	as	+
-	<math>2\~~left~~( ~~\frac{~~u}{2~~} \right~~)</math>	+
-	and use the formula for double angles (so as to end up with a right-hand side which has	+
-	<math>~~\frac{~~u}{2}</math>	+
-	as its argument)	+
-		+
-		+
-	<math>\frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}</math>	+
-		+

-	~~Writing the~~	+	{{Displayed math\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}
-	<math>\~~text~~{1}~~</math>~~	+
-	~~in the denominator as~~	+
-	~~<math>~~\cos ^{2}\frac{u}{2}+\sin ^{2}\~~frac~~{u}{2}</math>	+
-	~~using the Pythagorean identity,~~	+

		+	Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{2\cos \~~frac~~{u}{2}\~~centerdot~~ \sin \~~frac~~{u}{2}}{1+\cos ^{2}\~~frac~~{u}{2}-\sin ^{2}\~~frac~~{u}{2}}=\frac{2\cos \~~frac~~{u}{2}\sin \~~frac~~{u}{2}}{\cos ^{2}\~~frac~~{u}{2}+\sin ^{2}\~~frac~~{u}{2}+\cos ^{2}\~~frac~~{u}{2}-\sin ^{2}\~~frac~~{u}{2}} \\	+	\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}
-	& =\frac{2\cos \~~frac~~{u}{2}\sin \~~frac~~{u}{2}}{2\cos ^{2}\~~frac~~{u}{2}}=\frac{\sin \~~frac~~{u}{2}}{\cos \~~frac~~{u}{2}}=\tan \frac{u}{2} \\	+	&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt]
-	\end{align}</math>	+	&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{2\cos^2\dfrac{u}{2}}\\[5pt]
		+	&= \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}}\\[8pt]
		+	&= \tan\frac{u}{2}\,\textrm{.}
		+	\end{align}</math>}}

Ian um 11:08, 30. Sep. 2008

2008-09-30T11:08:10Z

← Nächstältere Version		Version vom 11:08, 30. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	One could write
-	<~~center~~> ~~[[Image:4_3_8c~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>\tan \frac{u}{2}</math>
-	{{~~NAVCONTENT_STOP}~~}	+	as a quotient involving sine and cosine, and then continue with the formula for half-angles,
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:4_3_8c-~~2(2~~).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...</math>
		+
		+
		+	but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.
		+
		+	We write
		+	<math>u</math>
		+	as
		+	<math>2\left( \frac{u}{2} \right)</math>
		+	and use the formula for double angles (so as to end up with a right-hand side which has
		+	<math>\frac{u}{2}</math>
		+	as its argument)
		+
		+
		+	<math>\frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}</math>
		+
		+
		+	Writing the
		+	<math>\text{1}</math>
		+	in the denominator as
		+	<math>\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}</math>
		+	using the Pythagorean identity,
		+
		+
		+	<math>\begin{align}
		+	& \frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}=\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}} \\
		+	& =\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{2\cos ^{2}\frac{u}{2}}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\tan \frac{u}{2} \\
		+	\end{align}</math>

Tekbot: Lösning 4.3:8c moved to Solution 4.3:8c: Robot: moved page

2008-09-09T09:58:43Z

Lösning 4.3:8c moved to Solution 4.3:8c: Robot: moved page

← Nächstältere Version

Version vom 09:58, 9. Sep. 2008

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

2008-08-21T07:16:29Z

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

← Nächstältere Version		Version vom 07:16, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_8c-1(2).gif]] </center>	+	<center> [[Image:4_3_8c-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_8c-2(2).gif]] </center>	+	<center> [[Image:4_3_8c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_3_8c-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_3_8c-2(2).gif {{NAVCONTENT_STOP}}

2008-04-03T10:04:45Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_8c-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_8c-2(2).gif </center> {{NAVCONTENT_STOP}}

Neue Seite

{{NAVCONTENT_START}}
<center> [[Bild:4_3_8c-1(2).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:4_3_8c-2(2).gif]] </center>
{{NAVCONTENT_STOP}}