Lösung 4.4:8c - Versionsgeschichte

Markus.bez um 15:13, 19. Jun. 2009

Markus.bez — Fri, 19 Jun 2009 15:13:59 GMT

← Nächstältere Version		Version vom 15:13, 19. Jun. 2009
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	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
-	x &= n\pi\,,\\[5pt]	+	x &= n\pi\,\text{ und}\\[5pt]
-	x &= \frac{3\pi}{4}+n\pi\,,	+	x &= \frac{3\pi}{4}+n\pi\,.
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

Markus.bez um 15:11, 19. Jun. 2009

Markus.bez — Fri, 19 Jun 2009 15:11:13 GMT

← Nächstältere Version		Version vom 15:11, 19. Jun. 2009
Zeile 1:		Zeile 1:
-	Wir versuchen die Gleichung durch trigonometrische Identitäten umzuschreiben, sodass sie nur ~~einen~~ trigonometrische Funktion enthält. Durch den trigonometrischen Pythagoras erhalten wir	+	Wir versuchen, die Gleichung durch trigonometrische Identitäten umzuschreiben, sodass sie nur eine trigonometrische Funktion enthält. Durch den trigonometrischen Pythagoras erhalten wir

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

-	~~Und wir~~ können die Gleichung in nur tan x-Terme ~~Schreiben,~~	+	Wir können die Gleichung in nur <math>\tan x</math>-Terme schreiben:

	{{Abgesetzte Formel\|\|<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}}

Martin um 12:18, 7. Apr. 2009

Martin — Tue, 07 Apr 2009 12:18:04 GMT

← Nächstältere Version		Version vom 12:18, 7. Apr. 2009
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-	~~When we have a trigonometric equation which contains a mixture of different trigonometric functions~~, ~~a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions~~. ~~Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “1” in the numerator of the left-hand side with <math>\sin^2\!x + \cos^2\!x</math>~~	+	Wir versuchen die Gleichung durch trigonometrische Identitäten umzuschreiben, sodass sie nur einen trigonometrische Funktion enthält. Durch den trigonometrischen Pythagoras erhalten wir
-	~~using the Pythagorean identity. This means that the equation's left-hand side can be written as~~	+

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

-	~~and the expression is then completely expressed~~ in ~~terms of~~ tan x,	+	Und wir können die Gleichung in nur tan x-Terme Schreiben,

	{{Abgesetzte Formel\|\|<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>1 + \tan^2\!x = 1 - \tan x\,\textrm{.}</math>}}

-	~~If we substitute~~ <math>t=\tan x</math>, ~~we see that we have a quadratic equation in~~ ''t'', ~~which, after simplifying, becomes~~ <math>t^2+t=0</math> ~~and has roots~~ <math>t=0</math> ~~and~~ <math>t=-1</math>. ~~There are therefore two possible values for~~	+	Benennen wir <math>t=\tan x</math>, erhalten wir eine quadratische Gleichung für ''t'', die vereinfacht <math>t^2+t=0</math> ist. Diese Gleichung hat die Lösungen <math>t=0</math> und <math>t=-1</math>. Daher muss ''x'' entweder die Gleichung <math>\tan x=0</math> oder die Gleichung <math>\tan x=-1\,</math> erfüllen. Die erste Gleichung hat die Lösungen <math>x=n\pi</math> und die zweite die Lösungen <math>x=3\pi/4+n\pi\,</math>.
-	~~<math>\tan~~ x~~</math>,~~ <math>\tan x=0</math> or <math>\tan x=-1\,</math>. ~~The first equality is satisfied when~~ <math>x=n\pi</math> ~~for all integers ''n'', and the second when~~ <math>x=3\pi/4+n\pi\,</math>.	+

-	~~The complete solution of the equation is~~	+	Also erhalten wir zusammen die Lösungen

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 17:		Zeile 15:
	x &= \frac{3\pi}{4}+n\pi\,,		x &= \frac{3\pi}{4}+n\pi\,,
	\end{align}\right.</math>}}		\end{align}\right.</math>}}
-
-	where ''n'' is an arbitrary integer.

Tekbot: hat „Solution 4.4:8c“ nach „Lösung 4.4:8c“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 15:19:44 GMT

hat „Solution 4.4:8c“ nach „Lösung 4.4:8c“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 15:19, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 09:01:38 GMT

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

← Nächstältere Version		Version vom 09:01, 22. Okt. 2008
Zeile 2:		Zeile 2:
	using the Pythagorean identity. This means that the equation's left-hand side can be written as		using the Pythagorean identity. This means that the equation's left-hand side can be written as

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

	and the expression is then completely expressed in terms of tan x,		and the expression is then completely expressed in terms of tan x,

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

	If we substitute <math>t=\tan x</math>, we see that we have a quadratic equation in ''t'', which, after simplifying, becomes <math>t^2+t=0</math> and has roots <math>t=0</math> and <math>t=-1</math>. There are therefore two possible values for		If we substitute <math>t=\tan x</math>, we see that we have a quadratic equation in ''t'', which, after simplifying, becomes <math>t^2+t=0</math> and has roots <math>t=0</math> and <math>t=-1</math>. There are therefore two possible values for
Zeile 13:		Zeile 13:
	The complete solution of the equation is		The complete solution of the equation is

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
	x &= n\pi\,,\\[5pt]		x &= n\pi\,,\\[5pt]
	x &= \frac{3\pi}{4}+n\pi\,,		x &= \frac{3\pi}{4}+n\pi\,,

Tek um 08:27, 14. Okt. 2008

Tek — Tue, 14 Oct 2008 08:27:41 GMT

← Nächstältere Version		Version vom 08:27, 14. Okt. 2008
Zeile 1:		Zeile 1:
-	When we have a trigonometric equation which contains a mixture of different trigonometric functions, a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions. Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “	+	When we have a trigonometric equation which contains a mixture of different trigonometric functions, a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions. Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “1” in the numerator of the left-hand side with <math>\sin^2\!x + \cos^2\!x</math>
-	~~<math>\text{1}</math>~~	+
-	” in the numerator of the left-hand side with	+
-	<math>\~~text{~~sin}^{\~~text{2}}~~x+\~~text{~~cos}^{\~~text{2}}~~x~~\text{ }~~</math>	+
	using the Pythagorean identity. This means that the equation's left-hand side can be written as		using the Pythagorean identity. This means that the equation's left-hand side can be written as

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

	and the expression is then completely expressed in terms of tan x,		and the expression is then completely expressed in terms of tan x,

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

-	~~<math>1+\tan ^{2}x=1-\tan x</math>~~	+	If we substitute <math>t=\tan x</math>, we see that we have a quadratic equation in ''t'', which, after simplifying, becomes <math>t^2+t=0</math> and has roots <math>t=0</math> and <math>t=-1</math>. There are therefore two possible values for
-		+	<math>\tan x</math>, <math>\tan x=0</math> or <math>\tan x=-1\,</math>. The first equality is satisfied when <math>x=n\pi</math> for all integers ''n'', and the second when <math>x=3\pi/4+n\pi\,</math>.
-		+
-	If we substitute	+
-	<math>t=\tan x</math>	+
-	, we see that we have a ~~second-degree~~ equation in	+
-	~~<math>~~t~~</math>~~	+
-	, which, after simplifying, becomes	+
-	<math>t^~~{\text{~~2~~}}\text{ }~~+t=0</math>	+
-	and has roots	+
-	<math>t=0</math>	+
-	and	+
-	<math>t=-~~\text{~~1}</math>. There are therefore two possible values for	+
-	<math>\tan x</math>,	+
-	<math>\tan x=0</math>	+
-	~~tan x =0~~ or	+
-	<math>\tan x=-1</math>	+
-	The first equality is satisfied when	+
-	<math>x=n\pi </math>	+
-	for all integers	+
-	~~<math>~~n~~</math>~~, and the second when	+
-	<math>x=~~\frac{~~3\pi }{4}+n\pi </math>.	+

	The complete solution of the equation is		The complete solution of the equation is

-		+	{{Displayed math\|\|<math>\left\{\begin{align}
-	<math>\left\{ \begin{~~array}{*{35}l~~}	+	x &= n\pi\,,\\[5pt]
-	x=n\pi \\	+	x &= \frac{3\pi}{4}+n\pi\,,
-	x=\frac{3\pi }{4}+n\pi \\	+	\end{align}\right.</math>}}
-	\end{~~array~~} \right.</math>	+
-	(	+	where ''n'' is an arbitrary integer.
-	~~<math>~~n~~</math>~~	+
-	an arbitrary integer).	+

Ian um 08:14, 2. Okt. 2008

Ian — Thu, 02 Oct 2008 08:14:28 GMT

← Nächstältere Version		Version vom 08:14, 2. Okt. 2008
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-	{{~~NAVCONTENT_START~~}}	+	When we have a trigonometric equation which contains a mixture of different trigonometric functions, a useful strategy can be to rewrite the equation so that it is expressed in terms of just one of the functions. Sometimes, it is not easy to find a way to rewrite it, but in the present case a plausible way is to replace the “
-	<~~center~~> ~~[[Image:4_4_8c~~-1(2~~).gif]]~~ </~~center~~>	+	<math>\text{1}</math>
-	{~~{NAVCONTENT_STOP}~~}	+	” in the numerator of the left-hand side with
-	{{~~NAVCONTENT_START~~}}	+	<math>\text{sin}^{\text{2}}x+\text{cos}^{\text{2}}x\text{ }</math>
-	<~~center~~> ~~[[Image:4_4_8c~~-~~2(2)~~.~~gif]]~~ </~~center~~>	+	using the Pythagorean identity. This means that the equation's left-hand side can be written as
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\frac{1}{\cos ^{2}x}=\frac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=1+\frac{\sin ^{2}x}{\cos ^{2}x}=1+\tan ^{2}x</math>
		+
		+
		+	and the expression is then completely expressed in terms of tan x,
		+
		+
		+	<math>1+\tan ^{2}x=1-\tan x</math>
		+
		+
		+	If we substitute
		+	<math>t=\tan x</math>
		+	, we see that we have a second-degree equation in
		+	<math>t</math>
		+	, which, after simplifying, becomes
		+	<math>t^{\text{2}}\text{ }+t=0</math>
		+	and has roots
		+	<math>t=0</math>
		+	and
		+	<math>t=-\text{1}</math>. There are therefore two possible values for
		+	<math>\tan x</math>,
		+	<math>\tan x=0</math>
		+	tan x =0 or
		+	<math>\tan x=-1</math>
		+	The first equality is satisfied when
		+	<math>x=n\pi </math>
		+	for all integers
		+	<math>n</math>, and the second when
		+	<math>x=\frac{3\pi }{4}+n\pi </math>.
		+
		+	The complete solution of the equation is
		+
		+
		+	<math>\left\{ \begin{array}{*{35}l}
		+	x=n\pi \\
		+	x=\frac{3\pi }{4}+n\pi \\
		+	\end{array} \right.</math>
		+	(
		+	<math>n</math>
		+	an arbitrary integer).

Tekbot: Lösning 4.4:8c moved to Solution 4.4:8c: Robot: moved page

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

Lösning 4.4:8c moved to Solution 4.4:8c: Robot: moved page

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Version vom 10:09, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:21:49 GMT

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

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Oskar — Thu, 03 Apr 2008 12:48:37 GMT

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