Lösung 4.4:4 - Versionsgeschichte

Tek: Replaced figure with metapost figure

Tek — Tue, 25 Aug 2009 13:17:14 GMT

Replaced figure with metapost figure

← Nächstältere Version		Version vom 13:17, 25. Aug. 2009
Zeile 9:		Zeile 9:
	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}

-	~~[[Image~~:~~4_4_4~~.~~gif\|~~center]]	+	<center>{{:4.4.4 - Solution - Two unit circles with angles 110° and 250°, respectively}}</center>

	Die allgemeine Lösung ist		Die allgemeine Lösung ist

Markus.bez: Sprache und Formulierung

Markus.bez — Fri, 19 Jun 2009 14:31:26 GMT

Sprache und Formulierung

← Nächstältere Version		Version vom 14:31, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir finden zuerst die allgemeine Lösung der Gleichung, und ~~sehen danach~~ welche der ~~Winkeln~~ im Intervall zwischen <math>0^{\circ}</math> und <math>360^{\circ}\,</math> liegen.	+	Wir finden zuerst die allgemeine Lösung der Gleichung und prüfen anschließend, welche der Winkel im Intervall zwischen <math>0^{\circ}</math> und <math>360^{\circ}\,</math> liegen.

-	Wir betrachten zuerst den Ausdruck <math>2v+10^{\circ}</math>, und erhalten die Lösung	+	Wir betrachten zuerst den Ausdruck <math>2v+10^{\circ}</math> und erhalten die Lösung

	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}

-	Es gibt noch eine Lösung im Einheitskreis~~, nämlich im dritten Quadrant,~~ mit ~~denselben~~ Winkel zur negativen ''y''-Achse, wie der Winkel <math>110^{\circ}</math> zur positiven ''y''-Achse hat, ~~nur~~ mit ~~anderen~~ Vorzeichen. Daher erhalten wir die zweite Lösung	+	Es gibt noch eine weitere Lösung im Einheitskreis mit demselben Winkel zur negativen ''y''-Achse wie der Winkel <math>110^{\circ}</math> zur positiven ''y''-Achse hat, nämlich im dritten Quadranten, und zwar mit umgekehrtem Vorzeichen. Daher erhalten wir die zweite Lösung

	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}
Zeile 11:		Zeile 11:
	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]

-	~~und die~~ allgemeine Lösung ist	+	Die allgemeine Lösung ist

-	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{und}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{und}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,.\end{align}\right.</math>}}

-	Lösen wir ''w'' erhalten wir	+	Lösen wir nach ''w'' auf, erhalten wir

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{und}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{und}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}

-	~~für~~ verschiedene ''n'' erhalten wir unter ~~anderen~~ die Lösungen	+	Für verschiedene ''n'' erhalten wir unter anderem die Lösungen


Zeile 73:		Zeile 73:


-	Hieraus ~~sehen~~ wir die Lösungen die im Intervall von <math>0^{\circ}</math> bis <math>360^{\circ}</math> ~~sind~~:	+	Hieraus erkennen wir die Lösungen, die im Intervall von <math>0^{\circ}</math> bis <math>360^{\circ}</math> liegen:

	{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{und}\quad v=300^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{und}\quad v=300^{\circ}\,\textrm{.}</math>}}

Martin um 10:46, 7. Apr. 2009

Martin — Tue, 07 Apr 2009 10:46:14 GMT

← Nächstältere Version		Version vom 10:46, 7. Apr. 2009
Zeile 75:		Zeile 75:
	Hieraus sehen wir die Lösungen die im Intervall von <math>0^{\circ}</math> bis <math>360^{\circ}</math> sind:		Hieraus sehen wir die Lösungen die im Intervall von <math>0^{\circ}</math> bis <math>360^{\circ}</math> sind:

-	{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{~~and~~}\quad v=300^{\circ}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{und}\quad v=300^{\circ}\,\textrm{.}</math>}}

Martin um 10:45, 7. Apr. 2009

Martin — Tue, 07 Apr 2009 10:45:54 GMT

← Nächstältere Version		Version vom 10:45, 7. Apr. 2009
Zeile 5:		Zeile 5:
	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}

-	Es gibt noch eine Lösung im Einheitskreis, nämlich im dritten Quadrant, mit denselben Winkel zur negativen ''y''-Achse, wie	+	Es gibt noch eine Lösung im Einheitskreis, nämlich im dritten Quadrant, mit denselben Winkel zur negativen ''y''-Achse, wie der Winkel <math>110^{\circ}</math> zur positiven ''y''-Achse hat, nur mit anderen Vorzeichen. Daher erhalten wir die zweite Lösung
-	~~There is then a further solution which satisfies~~ <math>~~0^{\circ}\le 2v + 10^{\circ}\le 360~~^{\circ}</math>~~, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative~~ ''y''-~~axis as <math>100^{\circ}</math> makes with the positive ''y''-axis~~, i.~~e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>~~	+
-	~~with the negative ''y''-axis and consequently~~	+

	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}
Zeile 13:		Zeile 11:
	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]

-	~~Now it is easy to write down the general solution,~~	+	und die allgemeine Lösung ist

-	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{~~and~~}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{und}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}

-	~~and if we make~~ ''v'' ~~the subject, we get~~	+	Lösen wir ''w'' erhalten wir

-	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{~~and~~}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{und}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}

-	~~For different values of the integers~~ ''n''~~, we see that the corresponding solutions are:~~	+	für verschiedene ''n'' erhalten wir unter anderen die Lösungen


Zeile 75:		Zeile 73:


-	~~From the table, we see that the solutions that are between~~ <math>0^{\circ}</math> ~~and~~ <math>360^{\circ}</math> ~~are~~	+	Hieraus sehen wir die Lösungen die im Intervall von <math>0^{\circ}</math> bis <math>360^{\circ}</math> sind:

	{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}

Martin um 09:30, 7. Apr. 2009

Martin — Tue, 07 Apr 2009 09:30:11 GMT

← Nächstältere Version		Version vom 09:30, 7. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The idea is first to find the general solution to the equation and then to see which angles lie between~~ <math>0^{\circ}</math> ~~and~~ <math>360^{\circ}\,</math>.	+	Wir finden zuerst die allgemeine Lösung der Gleichung, und sehen danach welche der Winkeln im Intervall zwischen <math>0^{\circ}</math> und <math>360^{\circ}\,</math> liegen.

-	~~If we start by considering the expression~~ <math>2v+10^{\circ}</math> ~~as an unknown~~, ~~then we have a usual basic trigonometric equation. One solution which we can see directly is~~	+	Wir betrachten zuerst den Ausdruck <math>2v+10^{\circ}</math>, und erhalten die Lösung

	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}

		+	Es gibt noch eine Lösung im Einheitskreis, nämlich im dritten Quadrant, mit denselben Winkel zur negativen ''y''-Achse, wie
	There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>		There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
	with the negative ''y''-axis and consequently		with the negative ''y''-axis and consequently

Tekbot: hat „Solution 4.4:4“ nach „Lösung 4.4:4“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:15:44 GMT

hat „Solution 4.4:4“ nach „Lösung 4.4:4“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:15, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:59:38 GMT

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

Tek um 13:42, 13. Okt. 2008

Tek — Mon, 13 Oct 2008 13:42:02 GMT

← Nächstältere Version		Version vom 13:42, 13. Okt. 2008
Zeile 1:		Zeile 1:
-	The idea is first to find the general solution to the equation and then to see which angles lie between	+	The idea is first to find the general solution to the equation and then to see which angles lie between <math>0^{\circ}</math> and <math>360^{\circ}\,</math>.
-	<math>0^{\circ }</math>	+
-	and	+
-	<math>360^{\circ }</math>.	+

-	If we start by considering the expression	+	If we start by considering the expression <math>2v+10^{\circ}</math> as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
-	<math>~~\text{2}v~~+~~\text{1}0~~^{\circ ~~}\text{~~ }</math>	+
-	as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is	+

		+	{{Displayed math\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}

-	~~<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }</math>~~	+	There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
-		+	with the negative ''y''-axis and consequently
-		+
-	There is then a further solution which satisfies	+
-	<math>0^{\circ }\le ~~\text{2}v~~+~~\text{1}0~~^{\circ }\le ~~\text{36}0~~^{\circ }</math>, where	+
-	<math>~~\text{2}v~~+~~\text{1}0~~^{\circ ~~}\text{~~ }</math>	+
-	lies in the third quadrant and makes the same angle with the negative y-axis as	+
-	<math>~~\text{1}00~~^{\circ }</math>	+
-	makes with the positive	+
-	~~<math>~~y~~</math>~~	+
-	-axis, i.e.	+
-	<math>~~\text{2}v~~+~~\text{1}0~~^{\circ ~~}\text{~~ }</math>	+
-	makes an angle	+
-	<math>~~\text{11}0~~^{\circ }-~~\text{9}0~~^{\circ }=~~\text{2}0~~^{\circ ~~}~\text{~~ }</math>	+
-	with the negative	+
-	~~<math>~~y~~</math>~~	+
-	-axis and consequently	+
-		+
-		+
-	~~<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>~~	+
-		+

		+	{{Displayed math\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}

	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]
-
-
-	There is then a further solution which satisfies
-	<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
-	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-	lies in the third quadrant and makes the same angle with the negative y-axis as
-	<math>\text{1}00^{\circ }</math>
-	makes with the positive
-	<math>y</math>
-	-axis, i.e.
-	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-	makes an angle
-	<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
-	with the negative
-	<math>y</math>
-	-axis and consequently
-
-
-	<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
-
-
-	FIGURE1 FIGURE2

	Now it is easy to write down the general solution,		Now it is easy to write down the general solution,

		+	{{Displayed math\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{and}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}

-	~~<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ }</math>~~	+	and if we make ''v'' the subject, we get
-	~~and~~	+
-		+
-	~~<math>\text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }</math>~~	+
-		+
-		+
-	and if we make	+
-	~~<math>~~v~~</math>~~	+
-	the subject, we get	+
-		+
-		+
-	~~<math>v=50^{\circ }+n\centerdot 180^{\circ }</math>~~	+
-	~~and~~	+
-		+
-	~~<math>v=120^{\circ }+n\centerdot 180^{\circ }</math>~~	+
-	~~EQ6~~	+
-		+
-	~~For different values of the integers~~	+
-	~~<math>n</math>, we see that the corresponding solutions are:~~	+

		+	{{Displayed math\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{and}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}

-	~~<math>\begin{array}{*{35}l}~~	+	For different values of the integers ''n'', we see that the corresponding solutions are:
-	~~\cdots \cdots & \cdots \cdots & \cdots \cdots \\~~	+
-	n~~=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\~~	+
-	~~n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\~~	+
-	~~n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\~~	+
-	~~n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\~~	+
-	~~n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\~~	+
-	~~n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\~~	+
-	~~\cdots \cdots & \cdots \cdots & \cdots \cdots \\~~	+
-	~~\end{array}</math>~~	+


		+	{\| align="center"
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|-
		+	\|align="left"\|<math>n=-2:</math>
		+	\|width="20px"\|
		+	\|align="left"\|<math>v = 50^{\circ} - 2\cdot 180^{\circ} = -310^{\circ}</math>
		+	\|width="20px"\|
		+	\|align="left"\|<math>v = 120^{\circ } - 2\cdot 180^{\circ} = -240^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=-1:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} - 1\cdot 180^{\circ} = -130^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} - 1\cdot 180^{\circ} = -60^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=0:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 0\cdot 180^{\circ} = 50^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 0\cdot 180^{\circ} = 120^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=1:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 1\cdot 180^{\circ} = 230^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 1\cdot 180^{\circ} = 300^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=2:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 2\cdot 180^{\circ} = 410^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 2\cdot 180^{\circ} = 480^{\circ}</math>
		+	\|-
		+	\|align="left"\|<math>n=3:</math>
		+	\|\|
		+	\|align="left"\|<math>v = 50^{\circ} + 3\cdot 180^{\circ} = 590^{\circ}</math>
		+	\|\|
		+	\|align="left"\|<math>v = 120^{\circ} + 3\cdot 180^{\circ} = 660^{\circ}</math>
		+	\|-
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|\|
		+	\|align="center"\|<math>\cdots\cdots</math>
		+	\|}

-	From the table, we see that the solutions that are between
-	<math>0^{\circ }</math>
-	and
-	<math>360^{\circ }</math>
-	are

		+	From the table, we see that the solutions that are between <math>0^{\circ}</math> and <math>360^{\circ}</math> are

-	<math>v=50,\quad v=120^{\circ },\quad v=230^{\circ }~~</math>~~	+	{{Displayed math\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}
-	and	+
-	~~<math>~~v=300^{\circ }</math>	+

Ian um 10:20, 1. Okt. 2008

Ian — Wed, 01 Oct 2008 10:20:04 GMT

← Nächstältere Version		Version vom 10:20, 1. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	The idea is first to find the general solution to the equation and then to see which angles lie between
-	<~~center~~> ~~[[Image:4_4_4-~~1~~(3)~~.~~gif]]~~ </~~center~~>	+	<math>0^{\circ }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and
-	{{~~NAVCONTENT_START~~}}	+	<math>360^{\circ }</math>.
-	<~~center~~> ~~[[Image:4_4_4~~-~~2(3)~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	If we start by considering the expression
-	{{~~NAVCONTENT_START~~}}	+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
-	<~~center~~> ~~[[Image:4_4_4-3(3).gif]]~~ </~~center~~>	+	as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }</math>
		+
		+
		+	There is then a further solution which satisfies
		+	<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	lies in the third quadrant and makes the same angle with the negative y-axis as
		+	<math>\text{1}00^{\circ }</math>
		+	makes with the positive
		+	<math>y</math>
		+	-axis, i.e.
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	makes an angle
		+	<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
		+	with the negative
		+	<math>y</math>
		+	-axis and consequently
		+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
		+


	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]
		+
		+
		+	There is then a further solution which satisfies
		+	<math>0^{\circ }\le \text{2}v+\text{1}0^{\circ }\le \text{36}0^{\circ }</math>, where
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	lies in the third quadrant and makes the same angle with the negative y-axis as
		+	<math>\text{1}00^{\circ }</math>
		+	makes with the positive
		+	<math>y</math>
		+	-axis, i.e.
		+	<math>\text{2}v+\text{1}0^{\circ }\text{ }</math>
		+	makes an angle
		+	<math>\text{11}0^{\circ }-\text{9}0^{\circ }=\text{2}0^{\circ }~\text{ }</math>
		+	with the negative
		+	<math>y</math>
		+	-axis and consequently
		+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=270^{\circ }-20^{\circ }=250^{\circ }</math>
		+
		+
		+	FIGURE1 FIGURE2
		+
		+	Now it is easy to write down the general solution,
		+
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=110^{\circ }+n\centerdot 360^{\circ }</math>
		+	and
		+
		+	<math>\text{2}v+\text{1}0^{\circ }=250^{\circ }+n\centerdot 360^{\circ }</math>
		+
		+
		+	and if we make
		+	<math>v</math>
		+	the subject, we get
		+
		+
		+	<math>v=50^{\circ }+n\centerdot 180^{\circ }</math>
		+	and
		+
		+	<math>v=120^{\circ }+n\centerdot 180^{\circ }</math>
		+	EQ6
		+
		+	For different values of the integers
		+	<math>n</math>, we see that the corresponding solutions are:
		+
		+
		+	<math>\begin{array}{*{35}l}
		+	\cdots \cdots & \cdots \cdots & \cdots \cdots \\
		+	n=-2 & v=50^{\circ }-2\centerdot 180^{\circ }=-310^{\circ } & v=120^{\circ }-2\centerdot 180^{\circ }=-240^{\circ } \\
		+	n=-1 & v=50^{\circ }-1\centerdot 180^{\circ }=-130^{\circ } & v=120^{\circ }-1\centerdot 180^{\circ }=-60^{\circ } \\
		+	n=0 & v=50^{\circ }+0\centerdot 180^{\circ }=50^{\circ } & v=120^{\circ }+0\centerdot 180^{\circ }=120^{\circ } \\
		+	n=1 & v=50^{\circ }+1\centerdot 180^{\circ }=230^{\circ } & v=120^{\circ }+1\centerdot 180^{\circ }=300^{\circ } \\
		+	n=2 & v=50^{\circ }+2\centerdot 180^{\circ }=410^{\circ } & v=120^{\circ }+2\centerdot 180^{\circ }=480^{\circ } \\
		+	n=3 & v=50^{\circ }+3\centerdot 180^{\circ }=590^{\circ } & v=120^{\circ }+3\centerdot 180^{\circ }=660^{\circ } \\
		+	\cdots \cdots & \cdots \cdots & \cdots \cdots \\
		+	\end{array}</math>
		+
		+
		+
		+	From the table, we see that the solutions that are between
		+	<math>0^{\circ }</math>
		+	and
		+	<math>360^{\circ }</math>
		+	are
		+
		+
		+	<math>v=50,\quad v=120^{\circ },\quad v=230^{\circ }</math>
		+	and
		+	<math>v=300^{\circ }</math>

Tekbot: Lösning 4.4:4 moved to Solution 4.4:4: Robot: moved page

Tekbot — Tue, 09 Sep 2008 10:05:23 GMT

Lösning 4.4:4 moved to Solution 4.4:4: Robot: moved page

← Nächstältere Version

Version vom 10:05, 9. Sep. 2008

← Nächstältere Version		Version vom 08:59, 22. Okt. 2008
Zeile 3:		Zeile 3:
	If we start by considering the expression <math>2v+10^{\circ}</math> as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is		If we start by considering the expression <math>2v+10^{\circ}</math> as an unknown, then we have a usual basic trigonometric equation. One solution which we can see directly is

-	{{~~Displayed math~~\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 110^{\circ}\,\textrm{.}</math>}}

	There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>		There is then a further solution which satisfies <math>0^{\circ}\le 2v + 10^{\circ}\le 360^{\circ}</math>, where <math>2v+10^{\circ}</math> lies in the third quadrant and makes the same angle with the negative ''y''-axis as <math>100^{\circ}</math> makes with the positive ''y''-axis, i.e. <math>2v + 10^{\circ}</math> makes an angle <math>110^{\circ} - 90^{\circ} = 20^{\circ}</math>
	with the negative ''y''-axis and consequently		with the negative ''y''-axis and consequently

-	{{~~Displayed math~~\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>2v + 10^{\circ} = 270^{\circ} - 20^{\circ} = 250^{\circ}\,\textrm{.}</math>}}

	[[Image:4_4_4.gif\|center]]		[[Image:4_4_4.gif\|center]]
Zeile 14:		Zeile 14:
	Now it is easy to write down the general solution,		Now it is easy to write down the general solution,

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{and}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} 2v + 10^{\circ} &= 110^{\circ} + n\cdot 360^{\circ}\quad\text{and}\\[5pt] 2v + 10^{\circ} &= 250^{\circ} + n\cdot 360^{\circ}\,,\end{align}\right.</math>}}

	and if we make ''v'' the subject, we get		and if we make ''v'' the subject, we get

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{and}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align} v &= 50^{\circ} + n\cdot 180^{\circ}\quad\text{and}\\[5pt] v &= 120^{\circ} + n\cdot 180^{\circ}\end{align}\right.</math>}}

	For different values of the integers ''n'', we see that the corresponding solutions are:		For different values of the integers ''n'', we see that the corresponding solutions are:
Zeile 76:		Zeile 76:
	From the table, we see that the solutions that are between <math>0^{\circ}</math> and <math>360^{\circ}</math> are		From the table, we see that the solutions that are between <math>0^{\circ}</math> and <math>360^{\circ}</math> are

-	{{~~Displayed math~~\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>v = 50^{\circ},\quad v=120^{\circ },\quad v=230^{\circ}\quad\text{and}\quad v=300^{\circ}\,\textrm{.}</math>}}