Solution 4.2:4f

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 4.2:4f moved to Solution 4.2:4f: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
If we add
-
<center> [[Image:4_2_4f.gif]] </center>
+
<math>2\pi </math>
-
{{NAVCONTENT_STOP}}
+
to
 +
<math>-\frac{5\pi }{3}</math>, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle
 +
<math>-\frac{5\pi }{3}</math>
 +
and consequently has the same tangent value:
 +
 
 +
 
 +
<math>\begin{align}
 +
& \tan \left( -\frac{5\pi }{3} \right)=\tan \left( -\frac{5\pi }{3}+2\pi \right)=\tan \frac{\pi }{3} \\
 +
& =\frac{\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \\
 +
\end{align}</math>

Revision as of 13:27, 28 September 2008

If we add \displaystyle 2\pi to \displaystyle -\frac{5\pi }{3}, we get a new angle in the first quadrant which corresponds to the same point on the unit circle as the old angle \displaystyle -\frac{5\pi }{3} and consequently has the same tangent value:


\displaystyle \begin{align} & \tan \left( -\frac{5\pi }{3} \right)=\tan \left( -\frac{5\pi }{3}+2\pi \right)=\tan \frac{\pi }{3} \\ & =\frac{\sin \frac{\pi }{3}}{\cos \frac{\pi }{3}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3} \\ \end{align}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_4.2:4f"