Solution 4.3:4b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:10, 9 October 2008) (edit) (undo)
m
 
Line 1: Line 1:
If we once again use the Pythagorean identity we get
If we once again use the Pythagorean identity we get
 +
{{Displayed math||<math>\cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \sin v = \pm\sqrt{1-\cos^2 v}\,\textrm{.}</math>}}
-
<math>\cos ^{2}v+\sin ^{2}v=1\quad \Leftrightarrow \quad \sin v=\pm \sqrt{1-\cos ^{2}v}</math>
+
Because the angle ''v'' lies between <math>0</math> and <math>\pi</math>, <math>\sin v</math> is positive (an angle in the first and second quadrants has a positive ''y''-coordinate) and therefore
-
 
+
{{Displayed math||<math>\sin v = +\sqrt{1-\cos^2 v} = \sqrt{1-b^2}\,\textrm{.}</math>}}
-
Because the angle v lies between
+
-
<math>0</math>
+
-
and
+
-
<math>\pi </math>,
+
-
<math>\text{sin }v</math>
+
-
is positive (an angle in the first and second quadrants has a positive
+
-
<math>y</math>
+
-
-coordinate) and therefore
+
-
 
+
-
 
+
-
<math>\sin v=+\sqrt{1-\cos ^{2}v}=\sqrt{1-b^{2}}</math>
+

Current revision

If we once again use the Pythagorean identity we get

\displaystyle \cos^2 v + \sin^2 v = 1\qquad\Leftrightarrow\qquad \sin v = \pm\sqrt{1-\cos^2 v}\,\textrm{.}

Because the angle v lies between \displaystyle 0 and \displaystyle \pi, \displaystyle \sin v is positive (an angle in the first and second quadrants has a positive y-coordinate) and therefore

\displaystyle \sin v = +\sqrt{1-\cos^2 v} = \sqrt{1-b^2}\,\textrm{.}
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.3:4b"