Solution 4.3:4b - Revision history

Tek at 14:10, 9 October 2008

2008-10-09T14:10:43Z

←Older revision		Revision as of 14:10, 9 October 2008
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	If we once again use the Pythagorean identity we get		If we once again use the Pythagorean identity we get

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

-	<math>\~~cos ^{2}v+~~\sin ~~^{2}~~v~~=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}~~</math>	+	Because the angle ''v'' lies between <math>0</math> and <math>\pi</math>, <math>\sin v</math> is positive (an angle in the first and second quadrants has a positive ''y''-coordinate) and therefore

-		+	{{Displayed math\|\|<math>\sin v = +\sqrt{1-\cos^2 v} = \sqrt{1-b^2}\,\textrm{.}</math>}}
-	~~Because the angle v lies between~~	+
-	~~<math>0</math>~~	+
-	~~and~~	+
-	~~<math>\pi </math>,~~	+
-	~~<math>\text~~{~~sin }v</~~math>	+
-	~~is positive (an angle in the first and second quadrants has a positive~~	+
-	~~<math>y</math>~~	+
-	~~-coordinate) and therefore~~	+
-		+
-		+
-	<math>\sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}</math>	+

2008-09-29T11:42:14Z

←Older revision		Revision as of 11:42, 29 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we once again use the Pythagorean identity we get
-	<~~center~~> ~~[[Image:4_3_4b.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\cos ^{2}v+\sin ^{2}v=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}</math>
		+
		+
		+	Because the angle v lies between
		+	<math>0</math>
		+	and
		+	<math>\pi </math>,
		+	<math>\text{sin }v</math>
		+	is positive (an angle in the first and second quadrants has a positive
		+	<math>y</math>
		+	-coordinate) and therefore
		+
		+
		+	<math>\sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}</math>

2008-09-09T09:54:23Z

←Older revision

Revision as of 09:54, 9 September 2008

2008-08-21T07:14:19Z

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2008-04-03T09:47:43Z

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