Solution 4.2:3c
From Förberedande kurs i matematik 1
m  (Lösning 4.2:3c moved to Solution 4.2:3c: Robot: moved page)  | 
				|||
| Line 1: | Line 1: | ||
| - | {{  | + | |
| - | <  | + | We can add and subtract multiples of   | 
| - | {{  | + | <math>\text{2}\pi </math>  | 
| + | to or from the argument of the sine function without changing its value. The angle   | ||
| + | <math>\text{2}\pi </math>  | ||
| + | corresponds to a whole turn in a unit circle and the sine function returns to the same value every time the angle changes by a complete revolution.  | ||
| + | |||
| + | For example, if we can subtract sufficiently many   | ||
| + | <math>\text{2}\pi </math>  | ||
| + | s from   | ||
| + | <math>\text{9}\pi </math>, we will obtain a more manageable argument which lies between   | ||
| + | <math>0</math>  | ||
| + | and  | ||
| + | <math>\text{2}\pi </math>,  | ||
| + | |||
| + | |||
| + | <math>\text{sin 9}\pi =\text{sin}\left( 9\pi -2\pi -2\pi -2\pi -2\pi  \right)=\sin \pi </math>  | ||
| + | |||
| + | |||
| + | The line which makes an angle   | ||
| + | <math>\pi </math>  | ||
| + | with the positive part of the   | ||
| + | <math>x</math>  | ||
| + | -axis is the negative part of the   | ||
| + | <math>x</math>  | ||
| + | -axis  | ||
| + | and it cuts the unit circle at the point   | ||
| + | <math>\left( -1 \right.,\left. 0 \right)</math>, which is why we can see from the   | ||
| + | <math>y</math>  | ||
| + | -coordinate that      | ||
| + | <math>\text{sin 9}\pi =\text{sin }\pi =0</math>.  | ||
| + | |||
| + | |||
| + | |||
[[Image:4_2_3_c.gif|center]]  | [[Image:4_2_3_c.gif|center]]  | ||
Revision as of 11:57, 28 September 2008
We can add and subtract multiples of \displaystyle \text{2}\pi to or from the argument of the sine function without changing its value. The angle \displaystyle \text{2}\pi corresponds to a whole turn in a unit circle and the sine function returns to the same value every time the angle changes by a complete revolution.
For example, if we can subtract sufficiently many \displaystyle \text{2}\pi s from \displaystyle \text{9}\pi , we will obtain a more manageable argument which lies between \displaystyle 0 and \displaystyle \text{2}\pi ,
\displaystyle \text{sin 9}\pi =\text{sin}\left( 9\pi -2\pi -2\pi -2\pi -2\pi  \right)=\sin \pi 
The line which makes an angle 
\displaystyle \pi 
with the positive part of the 
\displaystyle x
-axis is the negative part of the 
\displaystyle x
-axis
and it cuts the unit circle at the point 
\displaystyle \left( -1 \right.,\left. 0 \right), which is why we can see from the 
\displaystyle y
-coordinate that    
\displaystyle \text{sin 9}\pi =\text{sin }\pi =0.

