lilsis76 2 years ago Determine all of the solutions in the interval 0< Theta < 360. cos(Q/2) = 1/2 Q is theta

1. lilsis76

@zepdrix

2. Raja99

120,360

3. anikay

oh my, beaten to the punch gj

4. lilsis76

haha okay so then hmmm

5. lilsis76

how would that be? @zepdrix i dont get it. i see cos Q/2 that would mean cos = 1/2 right? and if its positive it would be in the 1st and 4th Quadrant?

6. anikay

well, now that I think about it, I think raja might be wrong

7. anikay

it should have 4 solutions right?

8. lilsis76

im not sure, it just says to determine all solutions in the interval 0-->360

9. Raja99

yes!

10. zepdrix

Using Q? Oooo I like that :) I'm always having trouble finding a suitable letter for theta in text, that works quite well XD

11. anikay

^concur

12. lilsis76

lol ya, i got tired ofputting theata.

13. Raja99

30

14. lilsis76

okay is it 60 and 300 degrees?

15. Raja99

no tetha is 30

16. zepdrix

$\large \cos(\theta/2)=\frac{1}{2}$We need to solve for theta, which involves a little more than just finding a special angle. If it helps, think of the ENTIRE inside of cosine as an angle, like this,$\large \cos(\phi)=\frac{1}{2}$Where is phi =1/2? Which special angles?

17. anikay

look, as I solve it, it's $\theta=2\cos^{-1} (1/2)$

18. anikay

the thing is, that only gives you 120. but it's a good starting point

19. lilsis76

mmmmm... okay, well first. special angles is like 45-45-90, and 30-60-90, right?

20. zepdrix

The special angles are 0,30, 45, 60, 90 In radians 0, pi/6, pi/4, pi/3, pi/2 and so on... Remember those? :) We're not talking about special triangles :D Special angles on the unit circle.

21. lilsis76

oh....okay i gotcha now on special angles. lol im totaly flipped around in this class

22. lilsis76

okay then so..cosQ/2 = 1/2 right? so cos would be a positive

23. anikay

I think 120 is it.....

24. zepdrix

yes the cosine is producing a positive value, so we're concerned with angles in the 1st and 4th quadrant right? :O Cuz that's where cosine is positive.

25. lilsis76

yes :) 1st and 4th

26. anikay

so, if we find the reference angle, then what would the two angles be?

27. lilsis76

mmm....pi/3 and 5pi/6?

28. anikay

yes to the first, the last, not so much, that's in the second quadrant

29. anikay

or am I failing?

30. zepdrix

lol :) They should both either be /6 or /3 :D you should get similar angles in that sense. Also, the domain of theta was given to us in DEGREES. let's try to keep it in degrees :)

31. lilsis76

oh. haha i mean 5pi/3

32. anikay

With zepdrix, but you are right

33. lilsis76

so the second answer would be 5pi/3?

34. anikay

but, the problem is, we need that angle to be doubled to match our equation of theta/2

35. lilsis76

hmmm...ok okkay, so doubled from pi/3, so pi/3 and 5pi/6?

36. anikay

uh, lets throw this back to degrees shall we? pi/3 = 60 and 5pi/3 = 300

37. lilsis76

DANG! haha okay, so first we have 60 and u said to double it, so that would be 120, 120 would be.....that 5pi/6

38. anikay

no, 5pi/6 = 150 degrees

39. lilsis76

haha shoot what am i doing wrong

40. anikay

here's a tip. every pi/6 = 30 degrees

41. lilsis76

pi/6 pi/4 pi/3 pi/2 4pi/6div by 2 = 2pi/3 OH!.....haha okay so pi/3 and 2 pi/3???

42. anikay

just plug those into your original equation and see what you get :P

43. lilsis76

i dont have an equatioN haha says determine all of the solutions in the interval 0< Q<360 cos(Q/2) = 1/2 so the answers would be pi/3 and 2pi/3? right?

44. lilsis76

@anikay

45. lilsis76

@zepdrix is it pi/3 and 2pi/3?

46. zepdrix

If the problem had been,$\large \cos\left(\theta\right)=\frac{1}{2}$We would have concluded that,$\large \theta=\quad 60^o,\quad 300^o$ But our problem involved Q/2, not Q. Giving us,$\large \cos\left(\frac{\theta}{2}\right)=\frac{1}{2}$ $\frac{\theta}{2}=\quad 60^o,\quad 300^o$ From here, if we multiply both sides by 2, we can solve for Q.

47. anikay

yup^

48. zepdrix

I'm a little confused why you're messing with the radians :C They're a little more confusing to work with. The domain of Q was GIVEN to us in DEGREES. So let's work in degrees! :D

49. lilsis76

AH! so i was right?! the 60 and 300?!

50. zepdrix

$\large \theta=\quad 120^o, \quad 600^o$

51. anikay

and since we're going only to 360?

52. anikay

(radians are easier imo but degrees work k)

53. lilsis76

AH!!! im so lost haha

54. zepdrix

aw :D

55. anikay

its almost like you're my lilsis or somethin xD

56. lilsis76

lol,

57. lilsis76

okay so i get the 1/2 and that is in the first quad giving me 60 degrees right?

58. lilsis76

@anikay

59. anikay

yes, and the 300

60. lilsis76

okay, how can i figure out the 300?!

61. anikay

it's a reference angle of 60

62. lilsis76

yes it is a ref angle 60

63. lilsis76

WAIT!!! 300 cuz its cos 1.2, sin - sqrt3 /2 ?!

64. lilsis76

right?! dang i think im slow

65. anikay

uh

66. lilsis76

haha did i do it right?!

67. anikay

yes

68. lilsis76

YAY!!!! > . < THANK YOU SO MUCH!!! ugh....

69. anikay

however, you have to double those and since you only go to 360, 120 is your only answer

70. phi

sometimes sketching the curve is helpful |dw:1355239813847:dw|

71. anikay

agreed, but my calculator is dead xD