Lösung 4.3:6b - Versionsgeschichte

Tek: Replaced figures with metapost figures

Tek — Tue, 25 Aug 2009 08:49:32 GMT

Replaced figures with metapost figures

← Nächstältere Version		Version vom 08:49, 25. Aug. 2009
Zeile 1:		Zeile 1:
	Wir zeichnen den Winkel <math>v</math> auf den Einheitskreis, dessen ''y''-Koordinate <math>\sin v = 3/10</math> entspricht:		Wir zeichnen den Winkel <math>v</math> auf den Einheitskreis, dessen ''y''-Koordinate <math>\sin v = 3/10</math> entspricht:

-	~~[[Image~~:~~4_3_6_b1~~.~~gif\|~~center]]	+	<center>{{:4.3.6b - Solution - The unit circle with angle v}}</center>

	Wir zeichnen im zweiten Quadrant ein rechtwinkliges Dreieck mit der Hypotenuse 1 und der Höhe 3/10.		Wir zeichnen im zweiten Quadrant ein rechtwinkliges Dreieck mit der Hypotenuse 1 und der Höhe 3/10.

-	~~[[Image~~:~~4_3_6_b2~~.~~gif\|~~center]]	+	<center>{{:4.3.6b - Solution - The unit circle with angle v in the second quadrant and an auxiliary triangle}}</center>

	Die Breite erhalten wir durch das Gesetz des Pythagoras:		Die Breite erhalten wir durch das Gesetz des Pythagoras:

Markus.bez: Sprache und Formulierung

Markus.bez — Fri, 19 Jun 2009 12:50:56 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 12:50, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir zeichnen den Winkel <math>v</math> im Einheitskreis, ~~mit der~~ ''y''-Koordinate ~~entsprechend~~ <math>\sin v = 3/10</math>	+	Wir zeichnen den Winkel <math>v</math> auf den Einheitskreis, dessen ''y''-Koordinate <math>\sin v = 3/10</math> entspricht:

	[[Image:4_3_6_b1.gif\|center]]		[[Image:4_3_6_b1.gif\|center]]
Zeile 7:		Zeile 7:
	[[Image:4_3_6_b2.gif\|center]]		[[Image:4_3_6_b2.gif\|center]]

-	Die Breite erhalten wir durch das ~~Gesets~~ des Pythagoras,	+	Die Breite erhalten wir durch das Gesetz des Pythagoras:

	{{Abgesetzte Formel\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}		{{Abgesetzte Formel\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}

-	~~und wir~~ erhalten	+	Wir erhalten:

	{{Abgesetzte Formel\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}

-	Also ist die ''x''Koordinate des Winkels <math>-a</math>, und wir erhalten	+	Also ist die ''x''Koordinate des Winkels <math>-a</math> und wir erhalten

	{{Abgesetzte Formel\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}		{{Abgesetzte Formel\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}

Martin um 12:46, 5. Apr. 2009

Martin — Sun, 05 Apr 2009 12:46:23 GMT

← Nächstältere Version		Version vom 12:46, 5. Apr. 2009
Zeile 1:		Zeile 1:
-	~~We draw an angle~~ <math>v</math> ~~in the unit circle~~, ~~and the fact that <math>\sin v = 3/10</math> means that its~~ ''y''-~~coordinate equals~~ <math>3/10</math>.	+	Wir zeichnen den Winkel <math>v</math> im Einheitskreis, mit der ''y''-Koordinate entsprechend <math>\sin v = 3/10</math>

	[[Image:4_3_6_b1.gif\|center]]		[[Image:4_3_6_b1.gif\|center]]

-	~~With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of~~ 1 ~~and a vertical side of length~~ 3/10.	+	Wir zeichnen im zweiten Quadrant ein rechtwinkliges Dreieck mit der Hypotenuse 1 und der Höhe 3/10.

	[[Image:4_3_6_b2.gif\|center]]		[[Image:4_3_6_b2.gif\|center]]

-	~~We can determine the triangle's remaining side by using the Pythagorean theorem~~,	+	Die Breite erhalten wir durch das Gesets des Pythagoras,

	{{Abgesetzte Formel\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}		{{Abgesetzte Formel\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}

-	~~which gives that~~	+	und wir erhalten

	{{Abgesetzte Formel\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}

-	~~This means that the angle's~~ ''x''~~-coordinate is~~ <math>-a</math>, ~~i.e. we have~~	+	Also ist die ''x''Koordinate des Winkels <math>-a</math>, und wir erhalten

	{{Abgesetzte Formel\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}		{{Abgesetzte Formel\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}

-	~~and thus~~	+	Also ist

	{{Abgesetzte Formel\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}

Tekbot: hat „Solution 4.3:6b“ nach „Lösung 4.3:6b“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:07:04 GMT

hat „Solution 4.3:6b“ nach „Lösung 4.3:6b“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:07, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:56:18 GMT

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

← Nächstältere Version		Version vom 08:56, 22. Okt. 2008
Zeile 9:		Zeile 9:
	We can determine the triangle's remaining side by using the Pythagorean theorem,		We can determine the triangle's remaining side by using the Pythagorean theorem,

-	{{~~Displayed math~~\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}	+	{{Abgesetzte Formel\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}

	which gives that		which gives that

-	{{~~Displayed math~~\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}

	This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have		This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have

-	{{~~Displayed math~~\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}	+	{{Abgesetzte Formel\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}

	and thus		and thus

-	{{~~Displayed math~~\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}

Tek um 13:00, 10. Okt. 2008

Tek — Fri, 10 Oct 2008 13:00:46 GMT

← Nächstältere Version		Version vom 13:00, 10. Okt. 2008
Zeile 1:		Zeile 1:
-	We draw an angle	+	We draw an angle <math>v</math> in the unit circle, and the fact that <math>\sin v = 3/10</math> means that its ''y''-coordinate equals <math>3/10</math>.
-	<math>v</math>	+
-	in the unit circle, and the fact that	+
-	<math>\~~text{~~sin }v~~\text{ }~~=~~\frac{~~3}{10}</math>	+
-	means that its	+
-	~~<math>~~y~~</math>~~	+
-	-coordinate equals	+
-	<math>~~\frac{~~3}{10}</math>.	+

	[[Image:4_3_6_b1.gif\|center]]		[[Image:4_3_6_b1.gif\|center]]

-	With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of	+	With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of 1 and a vertical side of length 3/10.
-	~~<math>\text{~~1~~}</math>~~	+
-	and a vertical side of length	+
-	~~<math>\frac{~~3}{10~~}</math>~~.	+
-		+
-		+

	[[Image:4_3_6_b2.gif\|center]]		[[Image:4_3_6_b2.gif\|center]]

-	We can determine the triangle's remaining side by using ~~Pythagoras'~~ theorem,	+	We can determine the triangle's remaining side by using the Pythagorean theorem,
-		+
-		+
-	~~<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>~~	+

		+	{{Displayed math\|\|<math>a^2 + \Bigl(\frac{3}{10}\Bigr)^2 = 1^2</math>}}

	which gives that		which gives that

		+	{{Displayed math\|\|<math>a = \sqrt{1-\Bigl(\frac{3}{10}\Bigr)^2} = \sqrt{1-\frac{9}{100}} = \sqrt{\frac{91}{100}} = \frac{\sqrt{91}}{10}\,\textrm{.}</math>}}

-	<math>~~a=\sqrt{1~~-~~\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}~~</math>	+	This means that the angle's ''x''-coordinate is <math>-a</math>, i.e. we have

-	~~This means that the angle's~~	+	{{Displayed math\|\|<math>\cos v=-\frac{\sqrt{91}}{10}</math>}}
-	<math~~>x</math>~~	+
-	~~-coordinate is~~	+
-	~~<math>-a</math>, i.e. we have~~	+
-		+
-		+
-	<math>\cos v=-\frac{\sqrt{91}}{10}</math>	+

	and thus		and thus

-		+	{{Displayed math\|\|<math>\tan v = \frac{\sin v}{\cos v} = \frac{\dfrac{3}{10}}{-\dfrac{\sqrt{91}}{10}} = -\frac{3}{\sqrt{91}}\,\textrm{.}</math>}}
-	<math>\tan v=\frac{\sin v}{\cos v}=\frac{\~~frac~~{3}{10}}{-\~~frac~~{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}</math>	+

Ian um 09:15, 10. Okt. 2008

Ian — Fri, 10 Oct 2008 09:15:43 GMT

← Nächstältere Version		Version vom 09:15, 10. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	We draw an angle
-	<~~center~~> ~~[[Image:4_3_6b-1(2).gif]]~~ </~~center~~>	+	<math>v</math>
-	{{~~NAVCONTENT_STOP~~}}	+	in the unit circle, and the fact that
-	{~~{NAVCONTENT_START}~~}	+	<math>\text{sin }v\text{ }=\frac{3}{10}</math>
-	<~~center~~> ~~[[Image:4_3_6b-2(2).gif]]~~ </~~center~~>	+	means that its
-	{{~~NAVCONTENT_STOP}~~}	+	<math>y</math>
		+	-coordinate equals
		+	<math>\frac{3}{10}</math>.

	[[Image:4_3_6_b1.gif\|center]]		[[Image:4_3_6_b1.gif\|center]]
		+
		+	With the information that is given, we can define a right-angled triangle in the second quadrant which has a hypotenuse of
		+	<math>\text{1}</math>
		+	and a vertical side of length
		+	<math>\frac{3}{10}</math>.
		+
		+

	[[Image:4_3_6_b2.gif\|center]]		[[Image:4_3_6_b2.gif\|center]]
		+
		+	We can determine the triangle's remaining side by using Pythagoras' theorem,
		+
		+
		+	<math>a^{2}+\left( \frac{3}{10} \right)^{2}=1^{2}</math>
		+
		+
		+	which gives that
		+
		+
		+	<math>a=\sqrt{1-\left( \frac{3}{10} \right)^{2}}=\sqrt{1-\frac{9}{100}}=\sqrt{\frac{91}{100}}=\frac{\sqrt{91}}{10}</math>
		+
		+	This means that the angle's
		+	<math>x</math>
		+	-coordinate is
		+	<math>-a</math>, i.e. we have
		+
		+
		+	<math>\cos v=-\frac{\sqrt{91}}{10}</math>
		+
		+	and thus
		+
		+
		+	<math>\tan v=\frac{\sin v}{\cos v}=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}</math>

Tekbot: Lösning 4.3:6b moved to Solution 4.3:6b: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:56:43 GMT

Lösning 4.3:6b moved to Solution 4.3:6b: Robot: moved page

← Nächstältere Version

Version vom 09:56, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:15:29 GMT

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

← Nächstältere Version		Version vom 07:15, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_6b-1(2).gif]] </center>	+	<center> [[Image:4_3_6b-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_3_6b-2(2).gif]] </center>	+	<center> [[Image:4_3_6b-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[~~Bild~~:4_3_6_b1.gif\|center]]	+	[[Image:4_3_6_b1.gif\|center]]

-	[[~~Bild~~:4_3_6_b2.gif\|center]]	+	[[Image:4_3_6_b2.gif\|center]]

Madde um 15:42, 18. Aug. 2008

Madde — Mon, 18 Aug 2008 15:42:26 GMT

← Nächstältere Version		Version vom 15:42, 18. Aug. 2008
Zeile 6:		Zeile 6:
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

-	[[Bild:4_3_6_b1.gif]]	+	[[Bild:4_3_6_b1.gif\|center]]
-	[[Bild:4_3_6_b2.gif]]	+
		+	[[Bild:4_3_6_b2.gif\|center]]