Solution 4.3:3d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 4.3:3d moved to Solution 4.3:3d: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
The expression for the angle
-
<center> [[Image:4_3_3d.gif]] </center>
+
<math>{\pi }/{2}\;-v</math>
-
{{NAVCONTENT_STOP}}
+
differs from
 +
<math>{\pi }/{2}\;</math>
 +
by as much as
 +
<math>-v\text{ }</math>
 +
differs from
 +
<math>0</math>. This means that
 +
<math>{\pi }/{2}\;</math>
 +
makes the same angle with the positive
 +
<math>y</math>
 +
-axis as
 +
<math>-v\text{ }</math>
 +
makes with the positive
 +
<math>x</math>
 +
-axis.
 +
 
 +
 
[[Image:4_3_3_d.gif|center]]
[[Image:4_3_3_d.gif|center]]
 +
 +
Angle
 +
<math>v</math>
 +
angle
 +
<math>\pi -v</math>
 +
 +
 +
Therefore, the angle
 +
<math>{\pi }/{2}\;-v</math>
 +
has a
 +
<math>y</math>
 +
-coordinate which is equal to the
 +
<math>x</math>
 +
-coordinate for the angle
 +
<math>v</math>, i.e.
 +
 +
 +
<math>\sin \left( {\pi }/{2}\;-v \right)=\cos v</math>
 +
 +
 +
and from exercise c, we know that
 +
<math>\cos v=\sqrt{1-a^{2}}</math>
 +
 +
 +
 +
<math>\sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}</math>

Revision as of 11:11, 29 September 2008

The expression for the angle \displaystyle {\pi }/{2}\;-v differs from \displaystyle {\pi }/{2}\; by as much as \displaystyle -v\text{ } differs from \displaystyle 0. This means that \displaystyle {\pi }/{2}\; makes the same angle with the positive \displaystyle y -axis as \displaystyle -v\text{ } makes with the positive \displaystyle x -axis.


Angle \displaystyle v angle \displaystyle \pi -v


Therefore, the angle \displaystyle {\pi }/{2}\;-v has a \displaystyle y -coordinate which is equal to the \displaystyle x -coordinate for the angle \displaystyle v, i.e.


\displaystyle \sin \left( {\pi }/{2}\;-v \right)=\cos v


and from exercise c, we know that \displaystyle \cos v=\sqrt{1-a^{2}}


\displaystyle \sin \left( \frac{\pi }{2}-v \right)=\sqrt{1-a^{2}}

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_4.3:3d"