Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 4.3:3c

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (13:29, 9 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
-
With the help of the Pythagorean identity, we can express
+
With the help of the Pythagorean identity, we can express <math>\cos v</math> in terms of <math>\sin v</math>,
-
<math>\cos v</math>
+
-
in terms of
+
-
<math>\text{sin }v</math>,
+
 +
{{Displayed math||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \cos v = \pm\sqrt{1-\sin^2 v}\,\textrm{.}</math>}}
-
<math>\cos ^{2}v+\sin ^{2}v=1</math>
+
In addition, we know that the angle <math>v</math> lies between <math>-\pi/2</math>
 +
and <math>\pi/2</math>, i.e. either in the first or fourth quadrant, where angles always have a positive ''x''-coordinate (cosine value); thus, we can conclude that
-
 
+
{{Displayed math||<math>\cos v = \sqrt{1-\sin^2 v} = \sqrt{1-a^2}\,\textrm{.}</math>}}
-
In addition, we know that the angle
+
-
<math>v</math>
+
-
lies between
+
-
<math>-{\pi }/{2}\;</math>
+
-
and
+
-
<math>{\pi }/{2}\;</math>, i.e. either in the first or fourth quadrant, where angles always have a positive
+
-
<math>x</math>
+
-
-coordinate (cosine value); thus, we can conclude that
+
-
 
+
-
 
+
-
<math>\cos v=\sqrt{1-\text{sin}^{2}\text{ }v}=\sqrt{1-a^{2}}</math>
+

Current revision

With the help of the Pythagorean identity, we can express cosv in terms of sinv,

cos2v+sin2v=1cosv=1−sin2v. 

In addition, we know that the angle v lies between −2 and 2, i.e. either in the first or fourth quadrant, where angles always have a positive x-coordinate (cosine value); thus, we can conclude that

cosv=1−sin2v=1−a2. 
Retrieved from "http://wiki.math.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_4.3:3c"