## anonymous 2 years ago Find COS 23pi/6?

1. anonymous

Do you have a calculator?

2. anonymous

no

3. anonymous

π=180∘

4. anonymous

so I replace pi with 180?

5. anonymous

yes

6. anonymous

so it'd be 23*pi divided by 6?

7. anonymous

exactly (:

8. anonymous

thanks :)

9. anonymous

welcome (:

10. anonymous

Is 12.03 Right?

11. anonymous

cosine of a value cannot be larger than 1...

12. anonymous

13. anonymous

See the graph of cosine:|dw:1389756073553:dw|

14. anonymous

yes

15. anonymous

Maximum of a y=cos x function is 1. Minimum of a y=cosx function is -1

16. anonymous

so do I need to graph this?

17. anonymous

|dw:1389756066410:dw|

18. anonymous

So, you cannot have COS 23pi/6 = 12.03

19. anonymous

$\cos\frac{23\pi}{6} = \cos(3(2\pi) + \frac{5\pi}{6}) = \cos\frac{5\pi}{6} = ...?$

20. anonymous

Using the attached drawing where does 23/pi fall?

21. anonymous

Okay sir's let me tell you that. Cos23pi/6 -2pi will give you the same answer

22. anonymous

idk where 23/pi falls

23. anonymous

Subtract 2pi from it

24. anonymous

The reason you don't know where it falls is becomes you have to do more than one revolution around the circle

25. anonymous

so i subtract 2pi fom 6?

26. anonymous

2pi = 12pi/6..So 23pi/6-12pi/6 = 11pi/6

27. anonymous

Okay good.

28. anonymous

Are you familiar with the unit circle in trigonometry?

29. anonymous

Now if radians are bothering you we can convert it to degrees.

30. anonymous

do i need like a specific calculator for this?

31. anonymous

32. anonymous

You could plug this into your calculator, but if this comes up on a test and you can't use one, you will be screwed.

33. anonymous

11pi/6*180/pi

34. anonymous

Can you explain how to do it manually?

35. anonymous

That gives us 330 degrees

36. anonymous

Do you know where that is on the unit circle?

37. anonymous

38. anonymous

yes

39. anonymous

Okay. tell me the coordinates for it.

40. anonymous

sqrt 2/2, -sqrt 2/2

41. anonymous

Not quite.

42. anonymous

That would be 315 degrees.

43. anonymous

ummm sqrt 3/2 , -1/2

44. anonymous

yes. sqrt(3)/2, -1/2

45. anonymous

Okay. You should know that Cosx= x

46. anonymous

So what is the value for Cos330?

47. anonymous

I got -.991...

48. anonymous

Keep this exact. You just gave me the coordinates. and I just told you CosƟ=x

49. anonymous

so what is the exact answer?

50. anonymous

hint: (x,y)

51. anonymous

Use the coordinates you gave me.

52. anonymous

sqrt 3/2 , -1/2

53. anonymous

Yes and I told you cosƟ=x so what is cos330?

54. anonymous

I don't get it /.\ I'm sorry

55. anonymous

when I plug in Cos330 it gives me the decimal I gave you

56. anonymous

|dw:1389757103607:dw|

57. anonymous

Okay also cosƟ=A/H This is the unit circle and the hypotenuse will always be one so basically cosƟ=H what is Cos330?

58. anonymous

ygarcia what calculator are you using btw?

59. anonymous

I don't have a calculator

60. anonymous

Okay. Well just finish it. Cos330=A. What is A in that triangle? A for adjacent to theta.

61. anonymous

-1/2?

62. anonymous

|dw:1389757356882:dw|

63. anonymous

No. Adjacent means next to. So what is next to theta that isn't the hypotenuse. -1/2 is opposite.

64. anonymous

sqrt 3/2

65. anonymous

There you go.

66. anonymous

so the solution is sqrt 3/2?

67. anonymous

cos330=Sqrt(3)/2

68. anonymous

Yes

69. anonymous

Now let's see if you learned anything. Tell me the sin330 if you know that sinƟ=y

70. anonymous

-1/2

71. anonymous

Very good. Just use the coordinates you gave me for that degree value. (sqrt(3)/2, -1/2) (x,y) SinƟ=y and the y value for 330 is -1/2 so that's the answer.

72. anonymous

So Cos= x value and Sin= y value?

73. anonymous

In reference to the circle

74. anonymous

Yeah. Because in the unit circle the radius is one, that's why it's called a unit circle, one unit. Anytime you have something over 1 It's just whatever the numerator is.

75. anonymous

Okay, Thanks SO MUCH!!!! I don't fully get it yet but what you explained to me helped A LOT :)

76. anonymous

You're welcome.