Lösung 4.3:3c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 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>
+

Version vom 13:29, 9. Okt. 2008

With the help of the Pythagorean identity, we can express \displaystyle \cos v in terms of \displaystyle \sin v,

Vorlage:Displayed math

In addition, we know that the angle \displaystyle v lies between \displaystyle -\pi/2 and \displaystyle \pi/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

Vorlage:Displayed math

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.3:3c“