Solution 4.3:4e - Revision history

Tek at 14:20, 9 October 2008

2008-10-09T14:20:05Z

←Older revision		Revision as of 14:20, 9 October 2008
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	The addition formula for sine gives us that		The addition formula for sine gives us that

		+	{{Displayed math\|\|<math>\sin\Bigl(v+\frac{\pi}{4}\Bigr) = \sin v\cdot\cos\frac{\pi }{4} + \cos v\cdot\sin\frac{\pi}{4}\,\textrm{.}</math>}}

-	<math>\sin ~~\left(~~ v+\~~frac~~{\pi }{4~~} \right~~)=\sin ~~v\centerdot \cos \frac{~~\pi }{4}+\~~cos v~~\~~centerdot \sin \frac~~{~~\pi }{4~~}.</math>	+	Because we know from exercise b that <math>\sin v = \sqrt{1-b^2}</math> we use that
		+	<math>\cos (\pi/4) = \sin (\pi/4) = 1/\!\sqrt{2}</math> to obtain

-		+	{{Displayed math\|\|<math>\sin\Bigl(v+\frac{\pi }{4}\Bigr) = \sqrt{1-b^2}\cdot\frac{1}{\sqrt{2}} + b\cdot\frac{1}{\sqrt{2}}\,\textrm{.}</math>}}
-	~~Because we know from exercise b that~~	+
-	~~<math>\sin v=\sqrt~~{~~1-b^~~{~~2}}</~~math>	+
-	~~we use that~~	+
-	<math>~~\cos \frac{\pi }{4}=~~\sin \~~frac{\pi }{4}=\frac{1}{\sqrt{2}}</math>~~	+
-	~~to obtain~~	+
-		+
-	~~<math>\sin \left~~( v+\frac{\pi }{4} \~~right~~)=\sqrt{1-b^{2}}\~~centerdot~~ \frac{1}{\sqrt{2}}+b\~~centerdot~~ \frac{1}{\sqrt{2}}.</math>	+

2008-09-29T11:54:32Z

←Older revision		Revision as of 11:54, 29 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The addition formula for sine gives us that
-	<~~center~~> ~~[[Image:4_3_4e.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\sin \left( v+\frac{\pi }{4} \right)=\sin v\centerdot \cos \frac{\pi }{4}+\cos v\centerdot \sin \frac{\pi }{4}.</math>
		+
		+
		+	Because we know from exercise b that
		+	<math>\sin v=\sqrt{1-b^{2}}</math>
		+	we use that
		+	<math>\cos \frac{\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}</math>
		+	to obtain
		+
		+	<math>\sin \left( v+\frac{\pi }{4} \right)=\sqrt{1-b^{2}}\centerdot \frac{1}{\sqrt{2}}+b\centerdot \frac{1}{\sqrt{2}}.</math>

2008-09-09T09:55:23Z

←Older revision

Revision as of 09:55, 9 September 2008

2008-08-21T07:14:49Z

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2008-04-03T09:49:54Z

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