From Förberedande kurs i matematik 1

Revision as of 08:02, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_4f.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (10:52, 9 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we add <math>2\pi</math> to <math>-5\pi/3\,</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle <math>-5\pi/3</math> and consequently has the same tangent value,
-	<center> [[Bild:4_2_4f.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\begin{align}
		+	\tan\Bigl(-\frac{5\pi}{3}\Bigr)
		+	= \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr)
		+	= \tan\frac{\pi}{3}
		+	= \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}}
		+	= \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}}
		+	= \sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

If we add \displaystyle 2\pi to \displaystyle -5\pi/3\,, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle \displaystyle -5\pi/3 and consequently has the same tangent value,

\displaystyle \begin{align} \tan\Bigl(-\frac{5\pi}{3}\Bigr) = \tan\Bigl(-\frac{5\pi}{3}+2\pi\Bigr) = \tan\frac{\pi}{3} = \frac{\sin\dfrac{\pi}{3}}{\cos\dfrac{\pi}{3}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} = \sqrt{3}\,\textrm{.} \end{align}