Solution 4.3:3c - Revision history

Tek at 13:29, 9 October 2008

2008-10-09T13:29:15Z

←Older revision		Revision as of 13:29, 9 October 2008
Line 1:		Line 1:
-	With the help of the Pythagorean identity, we can express	+	With the help of the Pythagorean identity, we can express <math>\cos v</math> in terms of <math>\sin v</math>,
-	<math>\cos v</math>	+
-	in terms of	+
-	<math>\~~text{~~sin }v</math>,	+

		+	{{Displayed math\|\|<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.}</math>}}

-	<math>\~~cos ^{~~2~~}v+~~\~~sin ^{~~2~~}v=1~~</math>	+	In addition, we know that the angle <math>v</math> lies between <math>-\pi/2</math>
		+	and <math>\pi/2</math>, i.e. either in the first or fourth quadrant, where angles always have a positive ''x''-coordinate (cosine value); thus, we can conclude that

-		+	{{Displayed math\|\|<math>\cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.}</math>}}
-	~~In addition, we know that the angle~~	+
-	~~<math>v</math>~~	+
-	~~lies between~~	+
-	~~<math>-~~{~~\pi }/~~{~~2}\;</~~math>	+
-	~~and~~	+
-	~~<math>{\pi }/{2}\;</math>, i.e. either in the first or fourth quadrant, where angles always have a positive~~	+
-	~~<math>x</math>~~	+
-	~~-coordinate (cosine value); thus, we can conclude that~~	+
-		+
-		+
-	<math>\cos v=\sqrt{1-\~~text{~~sin}^{2~~}\text{ }~~v}=\sqrt{1-a^{2}}</math>	+

2008-09-29T10:58:14Z

←Older revision		Revision as of 10:58, 29 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	With the help of the Pythagorean identity, we can express
-	<~~center~~> ~~[[Image:4_3_3c~~.~~gif]]~~ </~~center~~>	+	<math>\cos v</math>
-	{{~~NAVCONTENT_STOP~~}}	+	in terms of
		+	<math>\text{sin }v</math>,
		+
		+
		+	<math>\cos ^{2}v+\sin ^{2}v=1</math>
		+
		+
		+	In addition, we know that the angle
		+	<math>v</math>
		+	lies between
		+	<math>-{\pi }/{2}\;</math>
		+	and
		+	<math>{\pi }/{2}\;</math>, i.e. either in the first or fourth quadrant, where angles always have a positive
		+	<math>x</math>
		+	-coordinate (cosine value); thus, we can conclude that
		+
		+
		+	<math>\cos v=\sqrt{1-\text{sin}^{2}\text{ }v}=\sqrt{1-a^{2}}</math>

2008-09-09T09:52:43Z

←Older revision

Revision as of 09:52, 9 September 2008

2008-08-21T07:13:29Z

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

2008-04-03T09:44:38Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_3c.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:4_3_3c.gif]] </center>
{{NAVCONTENT_STOP}}