## anonymous 5 years ago what is the principal value of cos^-1{cos 23pi/20}

1. anonymous

$\frac{17\pi}{20}$

2. anonymous

i think, let me check with a calculator

3. anonymous

yeh its correct

4. anonymous

its an interesting method to solve , Ill do a quick picture now on paint and hopefully try to explain

5. anonymous

ok..m waitin

6. anonymous

Well, not the best picture , but good enough . Now, first thing to notice is that the answer is not 23pi / 20 , because 23pi/20 is not in the range of cos^-1

7. anonymous

we must get the angle into the interval 0<x<pi

8. anonymous

and 23pi/20 is a bit more than pi

9. anonymous

hey can you tell me what is meant by principal value?? i dont know if i get the idea u knw

10. anonymous

however, the cos graph is symmetrical about the minimum points , that is cos(23pi /20) does equal the same value as cos (17pi /20) because they are both the same distance from the minimums

11. anonymous

principal value just means , "your answer must be in the range 0<x<pi" pretty much only then does cos^-1(cos(x)) = x

12. anonymous

if you really want to check your understanding then try $\sin^{-1} (\sin(\frac{23\pi}{20}))$

13. anonymous

its a very similar idea

14. anonymous

using translations of the graph , you get the angle into the range -pi/2 to pi/2 ( because the range of sin^-1 is -pi/2 to pi/2 )

15. anonymous

then apply sin^-1(sin(x)) = x

16. anonymous

hmm.. i understand what you are saying. i would like to know how you got the value 17pi/20? i know its at a distance pi/4 from 23pi/4. i wanna know if there is some other way to work it out.

17. anonymous

there is other ways of working it our but I dont remember them at this point in time

18. anonymous

hey is the prin. value for \sin^{-1} (\sin(\frac{23\pi}{20})) 13 pi/20?

19. anonymous

$\sin^{-1} (\sin(\frac{23\pi}{20}))$

20. anonymous

no

21. anonymous

draw a picture of a sin graph this time

22. anonymous

yea i did..

23. anonymous

my picture is not good, but you get the idea

24. anonymous

23pi / 20 , is a bit more than pi , yes?

25. anonymous

now, we need to get the angle 23pi/20 into the range -pi/2 <= <= pi/2

26. anonymous

okk...

27. anonymous

28. anonymous

isnt it at a distance pi/2 from it?

29. anonymous

now, I know the graph of sin is symmetric about pi/2

30. anonymous

so if I find the distance between 23pi/20 and pi/2 , then this is the distance I must go to the left and I will still have the equivalent value of the function

31. anonymous

yup..

32. anonymous

now, the distance between pi/2 and 23pi/ 20 = (23/20 - 1/2) pi = 13pi/20

33. anonymous

this means that if I go 13pi/20 units left from pi/2 , then I would be at a point that is a reflection in the line x=pi/2

34. anonymous

ie sin(23pi / 20 ) = sin ( pi/2 - 13pi/20 ) = sin ( -3pi/20 )

35. anonymous

now, since the angle of the sin function is in the range of the invese sin function

36. anonymous

I can apply sin^-1(sin(x) ) = x

37. anonymous

38. anonymous

oh..i understand now. thankyou so much=)

39. anonymous

which can be checked on the calculator , put cal in radian mode , type in " sin^-1(sin(23pi /20) )"

40. anonymous

with the brackets, and you will get something like -0.47 .... which is equal to -3pi/20

41. anonymous

ya..got it