From Förberedande kurs i matematik 1

Revision as of 08:00, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_2_4c.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:41, 9 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	In exercise 4.2:3e, we studied the angle <math>3\pi/4</math> and found that
-	<center> [[Bild:4_2_4c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}</math>}}
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		+	Because <math>\tan x</math> is defined as <math>\frac{\sin x}{\cos x}</math>, we get immediately that
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		+	{{Displayed math||<math>\tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}</math>}}

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Current revision

In exercise 4.2:3e, we studied the angle \displaystyle 3\pi/4 and found that

\displaystyle \cos\frac{3\pi }{4} = -\frac{1}{\sqrt{2}}\qquad\text{and}\qquad\sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}\,\textrm{.}

Because \displaystyle \tan x is defined as \displaystyle \frac{\sin x}{\cos x}, we get immediately that

\displaystyle \tan\frac{3\pi}{4} = \frac{\sin\dfrac{3\pi}{4}}{\cos \dfrac{3\pi}{4}} = \frac{\dfrac{1}{\sqrt{2}}}{-\dfrac{1}{\sqrt{2}}} = -1\,\textrm{.}