Lösung 4.3:8a - Versionsgeschichte

Markus.bez um 13:07, 19. Jun. 2009

2009-06-19T13:07:59Z

← Nächstältere Version		Version vom 13:07, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir schreiben <math>\tan v</math> ~~wie~~ <math>\frac{\sin v}{\cos v}</math>, ~~und~~ also ist	+	Wir schreiben <math>\tan v</math> als <math>\frac{\sin v}{\cos v}</math>, also ist

	{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}

-	Durch ~~das Gesetz~~ des Pythagoras erhalten wir	+	Durch den Satz des Pythagoras erhalten wir

-	{{Abgesetzte Formel\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos^2\!v + \sin^2\!v = 1\,.</math>}}

-	~~wir~~ schreiben <math>\cos^2\!v</math> wie <math>1 - \sin^2\!v</math>, und erhalten ~~dadurch~~	+	Wir schreiben <math>\cos^2\!v</math> wie <math>1 - \sin^2\!v</math> und erhalten damit

	{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

2009-04-05T13:13:44Z

← Nächstältere Version		Version vom 13:13, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We rewrite~~ <math>\tan v</math> ~~on the left-hand side as~~ <math>\frac{\sin v}{\cos v}</math>, ~~so that~~	+	Wir schreiben <math>\tan v</math> wie <math>\frac{\sin v}{\cos v}</math>, und also ist

	{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}

-	~~If we then use the Pythagorean identity~~	+	Durch das Gesetz des Pythagoras erhalten wir

	{{Abgesetzte Formel\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}		{{Abgesetzte Formel\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}

-	~~and rewrite~~ <math>\cos^2\!v</math> ~~in the denominator as~~ <math>1 - \sin^2\!v</math>, ~~we get what we are looking for on the right-hand side. The whole calculation is~~	+	wir schreiben <math>\cos^2\!v</math> wie <math>1 - \sin^2\!v</math>, und erhalten dadurch

	{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

2008-10-22T15:08:24Z

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

← Nächstältere Version

Version vom 15:08, 22. Okt. 2008

2008-10-22T08:56:58Z

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

← Nächstältere Version		Version vom 08:56, 22. Okt. 2008
Zeile 1:		Zeile 1:
	We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that		We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that

-	{{~~Displayed math~~\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}

	If we then use the Pythagorean identity		If we then use the Pythagorean identity

-	{{~~Displayed math~~\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}

	and rewrite <math>\cos^2\!v</math> in the denominator as <math>1 - \sin^2\!v</math>, we get what we are looking for on the right-hand side. The whole calculation is		and rewrite <math>\cos^2\!v</math> in the denominator as <math>1 - \sin^2\!v</math>, we get what we are looking for on the right-hand side. The whole calculation is

-	{{~~Displayed math~~\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

2008-10-10T08:16:57Z

← Nächstältere Version		Version vom 08:16, 10. Okt. 2008
Zeile 1:		Zeile 1:
-		+	We rewrite <math>\tan v</math> on the left-hand side as <math>\frac{\sin v}{\cos v}</math>, so that
-		+
-	We rewrite	+
-	<math>\~~text{~~tan }v</math>	+
-	on the left-hand side as	+
-	<math>\frac{\sin v}{\cos v}</math>, so that	+
-		+
-		+
-	~~<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>~~	+

		+	{{Displayed math\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v}\,\textrm{.}</math>}}

	If we then use the Pythagorean identity		If we then use the Pythagorean identity

		+	{{Displayed math\|\|<math>\cos^2\!v + \sin^2\!v = 1</math>}}

-	~~<math>\cos ^{2}v+\sin ^{2}v=1</math>~~	+	and rewrite <math>\cos^2\!v</math> in the denominator as <math>1 - \sin^2\!v</math>, we get what we are looking for on the right-hand side. The whole calculation is
-		+
-		+
-	and rewrite	+
-	<math>\~~text{~~cos}^~~{\text{~~2}}v</math>	+
-	in the denominator as	+
-	<math>~~\text{~~1}-\~~text{~~sin}^{\~~text{2}}~~v~~\text{ }~~</math>, we get what we are looking for on the right-hand side. The whole calculation is	+
-		+

-	<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>	+	{{Displayed math\|\|<math>\tan^2\!v = \frac{\sin^2\!v}{\cos^2\!v} = \frac{\sin^2\!v}{1-\sin^2\!v}\,\textrm{.}</math>}}

2008-09-30T10:37:25Z

← Nächstältere Version		Version vom 10:37, 30. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:4_3_8a.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	We rewrite
		+	<math>\text{tan }v</math>
		+	on the left-hand side as
		+	<math>\frac{\sin v}{\cos v}</math>, so that
		+
		+
		+	<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}</math>
		+
		+
		+	If we then use the Pythagorean identity
		+
		+
		+	<math>\cos ^{2}v+\sin ^{2}v=1</math>
		+
		+
		+	and rewrite
		+	<math>\text{cos}^{\text{2}}v</math>
		+	in the denominator as
		+	<math>\text{1}-\text{sin}^{\text{2}}v\text{ }</math>, we get what we are looking for on the right-hand side. The whole calculation is
		+
		+
		+	<math>\tan ^{2}v=\frac{\sin ^{2}v}{\cos ^{2}v}=\frac{\sin ^{2}v}{1-\sin ^{2}v}</math>

2008-09-09T09:58:03Z

← Nächstältere Version

Version vom 09:58, 9. Sep. 2008

2008-08-21T07:16:09Z

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

2008-04-03T10:54:46Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_8a.gif </center> {{NAVCONTENT_STOP}}

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