Solution 4.3:8c - Revision history

Tek at 08:38, 10 October 2008

Tek — Fri, 10 Oct 2008 08:38:31 GMT

←Older revision		Revision as of 08:38, 10 October 2008
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-	One could write	+	One could write <math>\tan\frac{u}{2}</math> as a quotient involving sine and cosine, and then continue with the formula for half-angles,
-	<math>\tan \frac{u}{2}</math>	+
-	as a quotient involving sine and cosine, and then continue with the formula for half-angles,	+
-		+
-		+
-	~~<math>\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...</math>~~	+

		+	{{Displayed math\|\|<math>\tan\frac{u}{2} = \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}} = \ldots</math>}}

	but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.		but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.

-	We write	+	We write <math>u</math> as <math>2\cdot(u/2)</math>and use the formula for double angles (so as to end up with a right-hand side which has <math>u/2</math> as its argument),
-	<math>u</math>	+
-	as	+
-	<math>2\~~left~~( ~~\frac{~~u}{2~~} \right~~)</math>	+
-	and use the formula for double angles (so as to end up with a right-hand side which has	+
-	<math>~~\frac{~~u}{2}</math>	+
-	as its argument)	+
-		+
-		+
-	<math>\frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}</math>	+
-		+

-	~~Writing the~~	+	{{Displayed math\|\|<math>\frac{\sin u}{1+\cos u} = \frac{\sin \Bigl(2\cdot\dfrac{u}{2}\Bigr)}{1+\cos\Bigl(2\cdot\dfrac{u}{2}\Bigr)} = \frac{2\cos\dfrac{u}{2}\cdot \sin\dfrac{u}{2}}{1+\cos^2\cfrac{u}{2}-\sin^2\cfrac{u}{2}}\,\textrm{.}</math>}}
-	<math>\~~text~~{1}~~</math>~~	+
-	~~in the denominator as~~	+
-	~~<math>~~\cos ^{2}\frac{u}{2}+\sin ^{2}\~~frac~~{u}{2}</math>	+
-	~~using the Pythagorean identity,~~	+

		+	Writing the 1 in the denominator as <math>\cos^2(u/2) + \sin^2(u/2)</math> using the Pythagorean identity,

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{2\cos \~~frac~~{u}{2}\~~centerdot~~ \sin \~~frac~~{u}{2}}{1+\cos ^{2}\~~frac~~{u}{2}-\sin ^{2}\~~frac~~{u}{2}}=\frac{2\cos \~~frac~~{u}{2}\sin \~~frac~~{u}{2}}{\cos ^{2}\~~frac~~{u}{2}+\sin ^{2}\~~frac~~{u}{2}+\cos ^{2}\~~frac~~{u}{2}-\sin ^{2}\~~frac~~{u}{2}} \\	+	\frac{2\cos\dfrac{u}{2}\cdot\sin\dfrac{u}{2}}{1+\cos^2\dfrac{u}{2}-\sin^2\dfrac{u}{2}}
-	& =\frac{2\cos \~~frac~~{u}{2}\sin \~~frac~~{u}{2}}{2\cos ^{2}\~~frac~~{u}{2}}=\frac{\sin \~~frac~~{u}{2}}{\cos \~~frac~~{u}{2}}=\tan \frac{u}{2} \\	+	&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{\cos^2\dfrac{u}{2} + \sin^2\dfrac{u}{2} + \cos^2\dfrac{u}{2} - \sin^2\dfrac{u}{2}}\\[8pt]
-	\end{align}</math>	+	&= \frac{2\cos\dfrac{u}{2}\sin\dfrac{u}{2}}{2\cos^2\dfrac{u}{2}}\\[5pt]
		+	&= \frac{\sin\dfrac{u}{2}}{\cos\dfrac{u}{2}}\\[8pt]
		+	&= \tan\frac{u}{2}\,\textrm{.}
		+	\end{align}</math>}}

Ian at 11:08, 30 September 2008

Ian — Tue, 30 Sep 2008 11:08:10 GMT

←Older revision		Revision as of 11:08, 30 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	One could write
-	<~~center~~> ~~[[Image:4_3_8c~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>\tan \frac{u}{2}</math>
-	{{~~NAVCONTENT_STOP}~~}	+	as a quotient involving sine and cosine, and then continue with the formula for half-angles,
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:4_3_8c-~~2(2~~).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\tan \frac{u}{2}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=...</math>
		+
		+
		+	but because this leads to square roots and difficulties with keeping a check on the correct sign in front of the roots, it is perhaps simpler instead to go backwards and work with the right-hand side.
		+
		+	We write
		+	<math>u</math>
		+	as
		+	<math>2\left( \frac{u}{2} \right)</math>
		+	and use the formula for double angles (so as to end up with a right-hand side which has
		+	<math>\frac{u}{2}</math>
		+	as its argument)
		+
		+
		+	<math>\frac{\sin u}{1+\cos u}=\frac{\sin \left( 2\centerdot \frac{u}{2} \right)}{1+\cos \left( 2\centerdot \frac{u}{2} \right)}=\frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}</math>
		+
		+
		+	Writing the
		+	<math>\text{1}</math>
		+	in the denominator as
		+	<math>\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}</math>
		+	using the Pythagorean identity,
		+
		+
		+	<math>\begin{align}
		+	& \frac{2\cos \frac{u}{2}\centerdot \sin \frac{u}{2}}{1+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}}=\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{\cos ^{2}\frac{u}{2}+\sin ^{2}\frac{u}{2}+\cos ^{2}\frac{u}{2}-\sin ^{2}\frac{u}{2}} \\
		+	& =\frac{2\cos \frac{u}{2}\sin \frac{u}{2}}{2\cos ^{2}\frac{u}{2}}=\frac{\sin \frac{u}{2}}{\cos \frac{u}{2}}=\tan \frac{u}{2} \\
		+	\end{align}</math>

Tekbot: Lösning 4.3:8c moved to Solution 4.3:8c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:58:43 GMT

Lösning 4.3:8c moved to Solution 4.3:8c: Robot: moved page

←Older revision

Revision as of 09:58, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:16:29 GMT

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

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Oskar — Thu, 03 Apr 2008 10:04:45 GMT

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