Solution 4.3:7b - Revision history

Tek at 08:11, 10 October 2008

Tek — Fri, 10 Oct 2008 08:11:10 GMT

←Older revision		Revision as of 08:11, 10 October 2008
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-	Using the addition formula, we rewrite	+	Using the addition formula, we rewrite <math>\sin (x+y)</math> as
-	<math>\~~text{~~sin~~}\left~~( x+y ~~\right~~)</math>	+
-	as	+

		+	{{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>	+	If we use the same solution procedure as in exercise a, we use the Pythagorean identity <math>\cos^2\!v + \sin^2\!v = 1</math> to express the unknown factors <math>\sin x</math> and <math>\sin y</math> in terms of <math>\cos x</math> and <math>\cos y</math>,

		+	{{Displayed math\|\|<math>\begin{align}
		+	\sin x &= \pm\sqrt{1-\cos^2\!x} = \pm\sqrt{1-\bigl(\tfrac{2}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{4}{25}} = \pm\frac{\sqrt{21}}{5}\,,\\[5pt]
		+	\sin y &= \pm\sqrt{1-\cos^2\!y} = \pm\sqrt{1-\bigl(\tfrac{3}{5}\bigr)^2} = \pm\sqrt{1-\tfrac{9}{25}} = \pm\frac{4}{5}\,\textrm{.}
		+	\end{align}</math>}}

-	~~If we use the same solution procedure as in exercise a, we use the Pythagorean identity~~	+	The angles ''x'' and ''y'' lie in the first quadrant and both <math>\sin x</math> and <math>\sin y</math> are therefore positive, i.e.
-		+
-	~~<math>\cos ^{2}v+\sin ^{2}v=1</math>~~	+
-	~~to express the unknown factors~~	+
-	~~<math>x\text{ }</math>~~	+
-	~~and~~	+
-	~~<math>y\text{ }</math>~~	+
-	~~in terms of~~	+
-	~~<math>\text{cos }x\text{ }</math>~~	+
-	~~and~~	+
-	~~<math>\text{cos }y</math>,~~	+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \sin x=\pm \sqrt{1-\text{cos}^{2}x}=\pm \sqrt{1-\left( \frac{2}{5} \right)^{2}}=\pm \sqrt{1-\frac{4}{25}}=\pm \frac{\sqrt{21}}{5}, \\~~	+
-	~~& \sin y=\pm \sqrt{1-\text{cos}^{2}y}=\pm \sqrt{1-\left( \frac{3}{5} \right)^{2}}=\pm \sqrt{1-\frac{9}{25}}=\pm \frac{4}{5} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	The angles	+
-	~~<math>~~x~~\text{ }</math>~~	+
-	and	+
-	~~<math>~~y~~\text{ }</math>~~	+
-	lie in the first quadrant and both	+
-	<math>\~~text{~~sin }x~~\text{ }~~</math>	+
-	and	+
-	<math>\~~text{~~sin }y~~\text{ }~~</math>	+
-	are therefore positive, i.e.	+
-		+
-		+
-	~~<math>\sin x=\frac{\sqrt{21}}{5}</math>~~	+
-	~~and~~	+
-	~~<math>\sin y=\frac{4}{5}</math>~~	+

		+	{{Displayed math\|\|<math>\sin x = \frac{\sqrt{21}}{5}\qquad\text{and}\qquad\sin y = \frac{4}{5}\,\textrm{.}</math>}}

	Thus, the answer is		Thus, the answer is

-		+	{{Displayed math\|\|<math>\sin (x+y) = \frac{\sqrt{21}}{5}\cdot\frac{3}{5} + \frac{2}{5}\cdot\frac{4}{5} = \frac{3\sqrt{21}+8}{25}\,\textrm{.}</math>}}
-	<math>\~~text{~~sin~~}\left~~( x+y ~~\right~~)=\frac{\sqrt{21}}{5}\~~centerdot~~ \frac{3}{5}+\frac{2}{5}\~~centerdot~~ \frac{4}{5}=\frac{3\sqrt{21}+8}{25}</math>	+

Ian at 10:29, 30 September 2008

Ian — Tue, 30 Sep 2008 10:29:05 GMT

←Older revision		Revision as of 10:29, 30 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Using the addition formula, we rewrite
-	<~~center~~> ~~[[Image:4_3_7b~~.~~gif]]~~ </~~center~~>	+	<math>\text{sin}\left( x+y \right)</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as
		+
		+
		+	<math>\text{sin}\left( x+y \right)=\sin x\centerdot \cos y+\cos x\centerdot \sin y</math>
		+
		+
		+	If we use the same solution procedure as in exercise a, we use the Pythagorean identity
		+
		+	<math>\cos ^{2}v+\sin ^{2}v=1</math>
		+	to express the unknown factors
		+	<math>x\text{ }</math>
		+	and
		+	<math>y\text{ }</math>
		+	in terms of
		+	<math>\text{cos }x\text{ }</math>
		+	and
		+	<math>\text{cos }y</math>,
		+
		+	<math>\begin{align}
		+	& \sin x=\pm \sqrt{1-\text{cos}^{2}x}=\pm \sqrt{1-\left( \frac{2}{5} \right)^{2}}=\pm \sqrt{1-\frac{4}{25}}=\pm \frac{\sqrt{21}}{5}, \\
		+	& \sin y=\pm \sqrt{1-\text{cos}^{2}y}=\pm \sqrt{1-\left( \frac{3}{5} \right)^{2}}=\pm \sqrt{1-\frac{9}{25}}=\pm \frac{4}{5} \\
		+	\end{align}</math>
		+
		+
		+	The angles
		+	<math>x\text{ }</math>
		+	and
		+	<math>y\text{ }</math>
		+	lie in the first quadrant and both
		+	<math>\text{sin }x\text{ }</math>
		+	and
		+	<math>\text{sin }y\text{ }</math>
		+	are therefore positive, i.e.
		+
		+
		+	<math>\sin x=\frac{\sqrt{21}}{5}</math>
		+	and
		+	<math>\sin y=\frac{4}{5}</math>
		+
		+
		+	Thus, the answer is
		+
		+
		+	<math>\text{sin}\left( x+y \right)=\frac{\sqrt{21}}{5}\centerdot \frac{3}{5}+\frac{2}{5}\centerdot \frac{4}{5}=\frac{3\sqrt{21}+8}{25}</math>

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

Tekbot — Tue, 09 Sep 2008 09:57:43 GMT

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

←Older revision

Revision as of 09:57, 9 September 2008

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Tekbot — Thu, 21 Aug 2008 07:15:59 GMT

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Oskar — Thu, 03 Apr 2008 10:07:53 GMT

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