Solution 4.3:3c
From Förberedande kurs i matematik 1
(Difference between revisions)
			  			                                                      
		          
			m  (Lösning 4.3:3c moved to Solution 4.3:3c: Robot: moved page)  | 
				|||
| Line 1: | Line 1: | ||
| - | {{  | + | With the help of the Pythagorean identity, we can express   | 
| - | <  | + | <math>\cos v</math>  | 
| - | {{  | + | in terms of   | 
| + | <math>\text{sin }v</math>,  | ||
| + | |||
| + | |||
| + | <math>\cos ^{2}v+\sin ^{2}v=1</math>  | ||
| + | |||
| + | |||
| + | In addition, we know that the angle   | ||
| + | <math>v</math>  | ||
| + | lies between   | ||
| + | <math>-{\pi }/{2}\;</math>  | ||
| + | and   | ||
| + | <math>{\pi }/{2}\;</math>, i.e. either in the first or fourth quadrant, where angles always have a positive   | ||
| + | <math>x</math>  | ||
| + | -coordinate (cosine value); thus, we can conclude that  | ||
| + | |||
| + | |||
| + | <math>\cos v=\sqrt{1-\text{sin}^{2}\text{ }v}=\sqrt{1-a^{2}}</math>  | ||
Revision as of 10:58, 29 September 2008
With the help of the Pythagorean identity, we can express \displaystyle \cos v in terms of \displaystyle \text{sin }v,
\displaystyle \cos ^{2}v+\sin ^{2}v=1
In addition, we know that the angle 
\displaystyle v
lies between 
\displaystyle -{\pi }/{2}\;
and 
\displaystyle {\pi }/{2}\;, i.e. either in the first or fourth quadrant, where angles always have a positive 
\displaystyle x
-coordinate (cosine value); thus, we can conclude that
\displaystyle \cos v=\sqrt{1-\text{sin}^{2}\text{ }v}=\sqrt{1-a^{2}}
