## anonymous one year ago Find the sine, cosine, and tangent of 360 degrees. @zzr0ck3r @SithsAndGiggles

1. anonymous

@Nnesha

2. Nnesha

you can use *unit circle * where cos is x-coordinate and sin - is y-coordinate and tan = sin over cos

3. Nnesha

without calculator you need to apply 45-45-90 triangle theorem or 30-60-90

4. jdoe0001

|dw:1438299380960:dw|

5. anonymous

ok im lost

6. jdoe0001

check your Unit Circle as Nnesha suggested

7. anonymous

ok. what do i look for

8. Nnesha

thats why just use unit circle instead 45-45-90 theorem

9. Nnesha

360 degree

10. Nnesha

look 't jdoe's drawing

11. anonymous

ok

12. Nnesha

do you have unit circle sheet ? if not then google it

13. anonymous

got one

14. anonymous

i look for 360 on the unit circle?

15. anonymous

there is none because 360 IS 0 and vice versa right?

16. Nnesha

0 doesn't mean *nothing* :=)

17. Nnesha

there is an order pair right (x,y) where x=cos y =sin

18. anonymous

yes. (1,0)

19. Nnesha

|dw:1438299090911:dw|

20. Nnesha

yes so sin 360 = ? and cos 360 = ?

21. anonymous

sin 360 = 0 cos 360 = 1

22. anonymous

and tan 360 = 0

23. anonymous

ok thx. what about this one. The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

24. anonymous

i find where the point 1 and -1 are at

25. Nnesha

you have to draw a right triangle

26. anonymous

ok

27. anonymous

28. Nnesha

wait a sec i have to refresh the page *lag is real*

29. Nnesha

given point is (1,-1) so cos =1 (positive ) sin = -1 ( negative ) so at which quadrant cos is positive but sin is negative remember the CAST rule ?

30. anonymous

yes

31. anonymous

32. Nnesha

yep right

33. Nnesha

|dw:1438300117250:dw| now you need to find hyp apply Pythagorean theorem and find 3rd side of right triangle

34. Nnesha

Pythagorean theorem a^2 +b^2 = c^2 where c is hypotenuse longest side of right triangle

35. Nnesha

ello ?

36. Nnesha

if you have question , let me know plz :=)

37. Nnesha

|dw:1438301856130:dw| a=1 b=-1 c= ? use that formula $\huge\rm a^2+b^2=c^2$ substitute a for 1 substitute b for -1 solve for c

38. anonymous

c = 0

39. anonymous

40. Nnesha

how is it c ?

41. Nnesha

nope

42. Nnesha

first of all take square of a and b

43. Nnesha

$\huge\rm 1^2 +(-1)^2=c^2$ solve for c

44. Nnesha

45. anonymous

1+ 1 =2

46. Nnesha

yes right so c^2=2 now take square root both sides bec you need c not c^2

47. anonymous

root2

48. Nnesha

yep right |dw:1438302263294:dw| so what is sin theta = ?

49. Nnesha

$\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$ remember this

50. anonymous

so sin theta = -1/root 2

51. Nnesha

yes right you can't leave the radical at the denominator so multiply both denoimnator and numerator by sqr root 2

52. anonymous

root 2 sin theta = 1?

53. anonymous

-1/2

54. Nnesha

correction:$\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$

55. anonymous

56. Nnesha

nope sin theta = -1 over square root 2 you can't have the radical at the denominator thats why multiply numerator and denominator by square root 2

57. anonymous

shoot this is hard

58. anonymous

idk i gtg

59. anonymous

thx tho

60. Nnesha

$\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$ multiply

61. Nnesha

alright i have to go,too we will continue later

62. anonymous

ok

63. Nnesha

:$\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }$ here multiply numerator by numerator and denominator by denominator$\frac{ -1 \times \sqrt{2} }{ \sqrt{2} \times \sqrt{2}}$ sqrt{2} times sqrt{2} = sqrt{4} which is equal to 2 you will get $\frac{ -\sqrt{2} }{ 2 }$ now at what degrree sin equal to -sqrt{2} over 2 ?

64. Nnesha

same way to find value of cos that's it!

65. Nnesha

if you have any question let me know tomorrow. :=)