From Förberedande kurs i matematik 1

Revision as of 10:01, 3 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:4_3_6a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:4_3_6a-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (07:35, 10 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	If we think of the angle v as an angle in the unit circle, then ''v'' lies in the fourth quadrant and has ''x''-coordinate 3/4.
-	<center> [[Bild:4_3_6a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	[[Image:4_3_6_a1.gif|center]]
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:4_3_6a-2(2).gif]] </center>	+	If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to 1 and an opposite side equal to 3/4.
-	{{NAVCONTENT_STOP}}	+
		+	[[Image:4_3_6_a2.gif|center]]
		+
		+	Using the Pythagorean theorem, it is possible to determine the remaining side from
		+
		+	{{Displayed math||<math>b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2</math>}}
		+
		+	which gives that
		+
		+	{{Displayed math||<math>b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
		+
		+	Because the angle ''v'' belongs to the fourth quadrant, its ''y''-coordinate is negative and is therefore equal to <math>-b</math>, i.e.
		+
		+	{{Displayed math||<math>\sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}</math>}}
		+
		+	Thus, we have directly that
		+
		+	{{Displayed math||<math>\tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}</math>}}

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Current revision

If we think of the angle v as an angle in the unit circle, then v lies in the fourth quadrant and has x-coordinate 3/4.

If we enlarge the fourth quadrant, we see that we can make a right-angled triangle with hypotenuse equal to 1 and an opposite side equal to 3/4.

Using the Pythagorean theorem, it is possible to determine the remaining side from

\displaystyle b^2 + \Bigl(\frac{3}{4}\Bigr)^2 = 1^2

which gives that

\displaystyle b = \sqrt{1-\Bigl(\frac{3}{4}\Bigr)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\,\textrm{.}

Because the angle v belongs to the fourth quadrant, its y-coordinate is negative and is therefore equal to \displaystyle -b, i.e.

\displaystyle \sin v=-\frac{\sqrt{7}}{4}\,\textrm{.}

Thus, we have directly that

\displaystyle \tan v = \frac{\sin v}{\cos v} = \frac{-\sqrt{7}/4}{3/4} = -\frac{\sqrt{7}}{3}\,\textrm{.}