From Förberedande kurs i matematik 1

Revision as of 12:25, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_4_2a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_4_2a-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (14:13, 10 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We draw a unit circle and mark those angles on the circle which have a ''y''-coordinate of <math>\sqrt{3}/2</math>, in order to see which solutions lie between
-	<center> [[Bild:4_4_2a-1(2).gif]] </center>	+	<math>0</math> and <math>2\pi</math>.
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+	[[Image:4_4_2_a.gif|center]]
-	<center> [[Bild:4_4_2a-2(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	In the first quadrant, we recognize <math>x = \pi/3</math> as the angle which has a sine value of <math>\sqrt{3}/2</math> and then we have the reflectionally symmetric solution <math>x = \pi - \pi/3 = 2\pi/3</math> in the second quadrant.
		+
		+	Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of <math>2\pi</math>
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		+	{{Displayed math||<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,,</math>}}
		+
		+	where ''n'' is an arbitrary integer.
		+
		+
		+	Note: When we write that the complete solution is given by
		+
		+	{{Displayed math||<math>x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,\textrm{,}</math>}}
		+
		+	this means that for every integer ''n'', we obtain a solution to the equation:
		+
		+	{{Displayed math||<math>\begin{array}{llll}
		+	&n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt]
		+	&n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt]
		+	&n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt]
		+	&n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt]
		+	&n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt]
		+	&\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots
		+	\end{array}</math>}}

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Current revision

We draw a unit circle and mark those angles on the circle which have a y-coordinate of \displaystyle \sqrt{3}/2, in order to see which solutions lie between \displaystyle 0 and \displaystyle 2\pi.

In the first quadrant, we recognize \displaystyle x = \pi/3 as the angle which has a sine value of \displaystyle \sqrt{3}/2 and then we have the reflectionally symmetric solution \displaystyle x = \pi - \pi/3 = 2\pi/3 in the second quadrant.

Each of those solutions returns to itself after every revolution, so that we obtain the complete solution if we add multiples of \displaystyle 2\pi

\displaystyle x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,,

where n is an arbitrary integer.

Note: When we write that the complete solution is given by

\displaystyle x = \frac{\pi}{3}+2n\pi\qquad\text{and}\qquad x = \frac{2\pi}{3}+2n\pi\,\textrm{,}

this means that for every integer n, we obtain a solution to the equation:

\displaystyle \begin{array}{llll} &n=0:\quad &x=\frac{\pi}{3}\quad &x=\frac{2\pi }{3}\\[5pt] &n=-1:\quad &x=\frac{\pi}{3}+(-1)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-1)\cdot 2\pi\\[5pt] &n=1:\quad &x=\frac{\pi}{3}+1\cdot 2\pi\quad &x=\frac{2\pi}{3}+1\cdot 2\pi\\[5pt] &n=-2:\quad &x=\frac{\pi}{3}+(-2)\cdot 2\pi\quad &x=\frac{2\pi}{3}+(-2)\cdot 2\pi\\[5pt] &n=2:\quad &x=\frac{\pi}{3}+2\cdot 2\pi\quad &x=\frac{2\pi}{3}+2\cdot 2\pi\\[5pt] &\phantom{n}\vdots &\phantom{x}\vdots &\phantom{x}\vdots \end{array}