Solution 4.3:7a - Revision history

Tek at 07:59, 10 October 2008

Tek — Fri, 10 Oct 2008 07:59:37 GMT

←Older revision		Revision as of 07:59, 10 October 2008
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-	We can write the expression	+	We can write the expression <math>\sin (x+y)</math> in terms of <math>\sin x</math>, <math>\cos x</math>, <math>\sin y</math> and <math>\cos y</math> if we use the addition formula for sine,
-	<math>\~~text{~~sin~~}\left~~( x+y ~~\right~~)</math>	+
-	in terms of	+
-	<math>\~~text{~~sin }x</math>,	+
-	<math>\~~text{~~cos }x</math>,	+
-	<math>\~~text{~~sin }y</math>	+
-	and	+
-	<math>\~~text{~~cos }y</math>	+
-	if we use the addition formula for sine,	+

		+	{{Displayed math\|\|<math>\sin (x+y) = \sin x\cdot \cos y + \cos x\cdot \sin y\,\textrm{.}</math>}}

-	<math>\~~text{sin}\left(~~ x~~+y \right)=\sin x\centerdot~~ \cos y+\~~cos~~ x~~\centerdot~~ \sin y</math>	+	In turn, it is possible to express the factors <math>\cos x</math> and <math>\cos y</math> in terms of <math>\sin x</math> and <math>\sin y</math> by using the Pythagorean identity,

		+	{{Displayed math\|\|<math>\begin{align}
		+	\cos x &= \pm \sqrt{1-\sin^2\!x} = \pm \sqrt{1-(2/3)^2} = \pm\frac{\sqrt{5}}{3}\,,\\[5pt]
		+	\cos y &= \pm \sqrt{1-\sin^2\!y} = \pm \sqrt{1-(1/3)^{2}} = \pm \frac{2\sqrt{2}}{3}\,\textrm{.}
		+	\end{align}</math>}}

-	~~In turn, it is possible to express the factors~~	+	Because ''x'' and ''y'' are angles in the first quadrant, <math>\cos x</math> and <math>\cos y</math> are positive, so we in fact have
-	~~<math>\text{cos }~~x~~</math>~~	+
-	and	+
-	~~<math>\text{cos }~~y~~</math>~~	+
-	in ~~terms of~~	+
-	<math>\~~text{sin }~~x</math>	+
-	and	+
-	<math>\~~text{sin }~~y</math>	+
-	~~by using the Pythagorean identity~~,	+

		+	{{Displayed math\|\|<math>\cos x = \frac{\sqrt{5}}{3}\qquad\text{and}\qquad\cos y = \frac{2\sqrt{2}}{3}\,\textrm{.}</math>}}

-	<math>\begin{align}
-	& \cos x=\pm \sqrt{1-\text{sin}^{2}x}=\pm \sqrt{1-\left( {2}/{3}\; \right)^{2}}=\pm \frac{\sqrt{5}}{3} \\
-	& \cos y=\pm \sqrt{1-\text{sin}^{2}y}=\pm \sqrt{1-\left( {1}/{3}\; \right)^{2}}=\pm \frac{2\sqrt{2}}{3} \\
-	\end{align}</math>
-
-
-
-	Because
-	<math>x</math>
-	and
-	<math>y</math>
-	are angles in the first quadrant,
-	<math>\text{cos }x</math>
-	and
-	<math>\text{cos }y</math>
-	are positive, so we in fact have
-
-
-	<math>\cos x=\frac{\sqrt{5}}{3}</math>
-	and
-	<math>\cos y=\frac{2\sqrt{2}}{3}</math>
-
-
	Finally, we obtain		Finally, we obtain

-		+	{{Displayed math\|\|<math>\sin (x+y) = \frac{2}{3}\cdot \frac{2\sqrt{2}}{3} + \frac{\sqrt{5}}{3}\cdot \frac{1}{3} = \frac{4\sqrt{2} + \sqrt{5}}{9}\,\textrm{.}</math>}}
-	<math>\sin ~~\left~~( x+y ~~\right~~)=\frac{2}{3}\~~centerdot~~ \frac{2\sqrt{2}}{3}+\frac{\sqrt{5}}{3}\~~centerdot~~ \frac{1}{3}=\frac{4\sqrt{2}+\sqrt{5}}{9}</math>	+

Ian at 10:14, 30 September 2008

Ian — Tue, 30 Sep 2008 10:14:14 GMT

←Older revision		Revision as of 10:14, 30 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We can write the expression
-	<~~center~~> ~~[[Image:4_3_7a.gif]]~~ </~~center~~>	+	<math>\text{sin}\left( x+y \right)</math>
-	{{~~NAVCONTENT_STOP~~}}	+	in terms of
		+	<math>\text{sin }x</math>,
		+	<math>\text{cos }x</math>,
		+	<math>\text{sin }y</math>
		+	and
		+	<math>\text{cos }y</math>
		+	if we use the addition formula for sine,
		+
		+
		+	<math>\text{sin}\left( x+y \right)=\sin x\centerdot \cos y+\cos x\centerdot \sin y</math>
		+
		+
		+	In turn, it is possible to express the factors
		+	<math>\text{cos }x</math>
		+	and
		+	<math>\text{cos }y</math>
		+	in terms of
		+	<math>\text{sin }x</math>
		+	and
		+	<math>\text{sin }y</math>
		+	by using the Pythagorean identity,
		+
		+
		+	<math>\begin{align}
		+	& \cos x=\pm \sqrt{1-\text{sin}^{2}x}=\pm \sqrt{1-\left( {2}/{3}\; \right)^{2}}=\pm \frac{\sqrt{5}}{3} \\
		+	& \cos y=\pm \sqrt{1-\text{sin}^{2}y}=\pm \sqrt{1-\left( {1}/{3}\; \right)^{2}}=\pm \frac{2\sqrt{2}}{3} \\
		+	\end{align}</math>
		+
		+
		+
		+	Because
		+	<math>x</math>
		+	and
		+	<math>y</math>
		+	are angles in the first quadrant,
		+	<math>\text{cos }x</math>
		+	and
		+	<math>\text{cos }y</math>
		+	are positive, so we in fact have
		+
		+
		+	<math>\cos x=\frac{\sqrt{5}}{3}</math>
		+	and
		+	<math>\cos y=\frac{2\sqrt{2}}{3}</math>
		+
		+
		+	Finally, we obtain
		+
		+
		+	<math>\sin \left( x+y \right)=\frac{2}{3}\centerdot \frac{2\sqrt{2}}{3}+\frac{\sqrt{5}}{3}\centerdot \frac{1}{3}=\frac{4\sqrt{2}+\sqrt{5}}{9}</math>

Tekbot: Lösning 4.3:7a moved to Solution 4.3:7a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:57:23 GMT

Lösning 4.3:7a moved to Solution 4.3:7a: Robot: moved page

←Older revision

Revision as of 09:57, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 07:15:49 GMT

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

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Oskar — Thu, 03 Apr 2008 10:07:09 GMT

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