Solution 4.2:4b
From Förberedande kurs i matematik 1
m  (Lösning 4.2:4b moved to Solution 4.2:4b: Robot: moved page)  | 
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| - | {{  | + | We start by subtracting   | 
| - | <  | + | <math>2\pi </math>  | 
| - | {{  | + | from  | 
| - | {{  | + | <math>\frac{11\pi }{3}</math>, so that we get an angle between   | 
| - | <  | + | <math>o</math>  | 
| - | {{  | + | and  | 
| + | <math>2\pi </math>. This doesn't change the cosine value  | ||
| + | |||
| + | |||
| + | <math>\cos \frac{11\pi }{3}=\cos \left( \frac{11\pi }{3}-2\pi  \right)=\cos \frac{5\pi }{3}</math>  | ||
| + | |||
| + | |||
| + | Then, by rewriting   | ||
| + | <math>\frac{5\pi }{3}</math>  | ||
| + | as a sum of   | ||
| + | <math>\pi </math>  | ||
| + | - and   | ||
| + | <math>\frac{\pi }{2}</math>  | ||
| + | -terms  | ||
| + | |||
| + | |||
| + | <math>\frac{5\pi }{3}=\frac{3\pi +\frac{3}{2}\pi +\frac{1}{2}\pi }{3}=\pi +\frac{\pi }{2}+\frac{\pi }{6}</math>  | ||
| + | |||
| + | we see that   | ||
| + | <math>\frac{5\pi }{3}</math>  | ||
| + | is an angle in the fourth quadrant which makes an angle   | ||
| + | <math>\frac{\pi }{6}</math>  | ||
| + | with the negative   | ||
| + | <math>y</math>  | ||
| + | -axis.  | ||
| + | |||
| + | |||
[[Image:4_2_4b1.gif]]  | [[Image:4_2_4b1.gif]]  | ||
| + | |||
| + | With the help of a triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle  | ||
| + | <math>\frac{5\pi }{3}</math>        .  | ||
| + | |||
| + | |||
[[Image:4_2_4_b2.gif]]  | [[Image:4_2_4_b2.gif]]  | ||
| + | |||
| + | The point has coordinates    | ||
| + | <math>\left( \frac{1}{2} \right.,\left. -\frac{\sqrt{3}}{2} \right)</math>  | ||
| + | and   | ||
| + | <math>\cos \frac{11\pi }{3}=\cos \frac{5\pi }{3}=\frac{1}{2}</math>.  | ||
Revision as of 13:05, 28 September 2008
We start by subtracting \displaystyle 2\pi from \displaystyle \frac{11\pi }{3}, so that we get an angle between \displaystyle o and \displaystyle 2\pi . This doesn't change the cosine value
\displaystyle \cos \frac{11\pi }{3}=\cos \left( \frac{11\pi }{3}-2\pi  \right)=\cos \frac{5\pi }{3}
Then, by rewriting 
\displaystyle \frac{5\pi }{3}
as a sum of 
\displaystyle \pi 
- and 
\displaystyle \frac{\pi }{2}
-terms
\displaystyle \frac{5\pi }{3}=\frac{3\pi +\frac{3}{2}\pi +\frac{1}{2}\pi }{3}=\pi +\frac{\pi }{2}+\frac{\pi }{6}
we see that \displaystyle \frac{5\pi }{3} is an angle in the fourth quadrant which makes an angle \displaystyle \frac{\pi }{6} with the negative \displaystyle y -axis.
With the help of a triangle and a little trigonometry, we can determine the coordinates for the point on a unit circle which corresponds to the angle \displaystyle \frac{5\pi }{3} .
The point has coordinates \displaystyle \left( \frac{1}{2} \right.,\left. -\frac{\sqrt{3}}{2} \right) and \displaystyle \cos \frac{11\pi }{3}=\cos \frac{5\pi }{3}=\frac{1}{2}.


