anonymous one year ago Use the unit circle to determine the value of sec(-12pi)

1. Nnesha

to make it easy add 2pi into -12pi ntill you get positive value and remember sec =1/cos

2. anonymous

i still dont get it

3. Nnesha

add 2pi $\huge\rm -12\pi +2\pi$ full circle is equal to 2pi that's why we have to add 2pi keep adding 2 untill u get positive number

4. Nnesha

$-12\pi +2\pi = -10$ still negative add 2pi again $-10\pi +2\pi = -8\pi$ add 2pi again

5. anonymous

ok i get 0pi

6. anonymous

so it is not an exact value then

7. Nnesha

$-12\pi +2\pi = -10$ still negative add 2pi again $-10\pi +2\pi = -8\pi$ add 2pi again $-8\pi +2\pi = -6\pi$ add 2pi $-6\pi +2\pi = -4\pi$ $-4\pi +2\pi = -2\pi$ add 2 pi again and let me know what you get

8. anonymous

i get 0pi

9. Nnesha

okay 0 is correct

10. Nnesha

not pi just 0

11. anonymous

so it is undefined

12. Nnesha

like i told you sec =1/cos look at the unit cricle what is cos at 0

13. anonymous

cos(0)=1

14. Nnesha

yep right so $\huge\rm sec =\frac{1}{cos} =sec =\frac{ 1 }{ 1 }$ replace cos by 1 = answer

15. Nnesha

don't forget the *one* at the numerator !

16. anonymous

if the answer is 0 the is it undefined or not

17. Nnesha

no

18. Nnesha

they are asking for value of sec(-12pi) in the unit circle radians is 0 to 2pi(-12pi is not in the unit circle t decrease the value we added 2pi ) degree 0 to 360

19. Nnesha

so in other words question is find the value fo sec(0)

20. Nnesha

sec(0) is same as sec(-12pi) both are equal

21. anonymous

no i am just asking in general, does the answer 0 mean undefined

22. Nnesha

ohh okay no well if the DENOMINATOR is 0 then YES!! answer is undefined

23. Nnesha

okay here is an example let's say they are asking csc(0) and csc equal to 1/sin and now sin (0) is 0 right

24. Nnesha

so when you replace sin by 0 your answer would be undefined because there is a zero at the denominator $\large \rm csc = \frac{ 1}{\sin} = \frac{ 1 }{ 0 }=\rm undefined$

25. Nnesha

$\huge\rm \frac{ 1 }{ 0 }= undefined ~~~~~~~~~~~~~\frac{ 0 }{ 1}=1$ and just simple 0 doesn't mean undefined

Find more explanations on OpenStudy