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anonymous

  • one year ago

Find the sine, cosine, and tangent of 360 degrees. @zzr0ck3r @SithsAndGiggles

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  1. anonymous
    • one year ago
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    @Nnesha

  2. Nnesha
    • one year ago
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    you can use *unit circle * where cos is x-coordinate and sin - is y-coordinate and tan = sin over cos

  3. Nnesha
    • one year ago
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    without calculator you need to apply 45-45-90 triangle theorem or 30-60-90

  4. jdoe0001
    • one year ago
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    |dw:1438299380960:dw|

  5. anonymous
    • one year ago
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    ok im lost

  6. jdoe0001
    • one year ago
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    check your Unit Circle as Nnesha suggested

  7. anonymous
    • one year ago
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    ok. what do i look for

  8. Nnesha
    • one year ago
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    thats why just use unit circle instead 45-45-90 theorem

  9. Nnesha
    • one year ago
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    360 degree

  10. Nnesha
    • one year ago
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    look 't jdoe's drawing

  11. anonymous
    • one year ago
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    ok

  12. Nnesha
    • one year ago
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    do you have unit circle sheet ? if not then google it

  13. anonymous
    • one year ago
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    got one

  14. anonymous
    • one year ago
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    i look for 360 on the unit circle?

  15. anonymous
    • one year ago
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    there is none because 360 IS 0 and vice versa right?

  16. Nnesha
    • one year ago
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    0 doesn't mean *nothing* :=)

  17. Nnesha
    • one year ago
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    there is an order pair right (x,y) where x=cos y =sin

  18. anonymous
    • one year ago
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    yes. (1,0)

  19. Nnesha
    • one year ago
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    |dw:1438299090911:dw|

  20. Nnesha
    • one year ago
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    yes so sin 360 = ? and cos 360 = ?

  21. anonymous
    • one year ago
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    sin 360 = 0 cos 360 = 1

  22. anonymous
    • one year ago
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    and tan 360 = 0

  23. anonymous
    • one year ago
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    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
    • one year ago
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    i find where the point 1 and -1 are at

  25. Nnesha
    • one year ago
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    you have to draw a right triangle

  26. anonymous
    • one year ago
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    ok

  27. anonymous
    • one year ago
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    from quadrant 1?

  28. Nnesha
    • one year ago
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    wait a sec i have to refresh the page *lag is real*

  29. Nnesha
    • one year ago
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    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
    • one year ago
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    yes

  31. anonymous
    • one year ago
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    4th quadrent

  32. Nnesha
    • one year ago
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    yep right

  33. Nnesha
    • one year ago
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    |dw:1438300117250:dw| now you need to find hyp apply Pythagorean theorem and find 3rd side of right triangle

  34. Nnesha
    • one year ago
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    Pythagorean theorem a^2 +b^2 = c^2 where c is hypotenuse longest side of right triangle

  35. Nnesha
    • one year ago
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    ello ?

  36. Nnesha
    • one year ago
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    if you have question , let me know plz :=)

  37. Nnesha
    • one year ago
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    |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
    • one year ago
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    c = 0

  39. anonymous
    • one year ago
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    thats the answer???

  40. Nnesha
    • one year ago
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    how is it c ?

  41. Nnesha
    • one year ago
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    nope

  42. Nnesha
    • one year ago
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    first of all take square of a and b

  43. Nnesha
    • one year ago
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    \[\huge\rm 1^2 +(-1)^2=c^2\] solve for c

  44. Nnesha
    • one year ago
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    and then add them

  45. anonymous
    • one year ago
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    1+ 1 =2

  46. Nnesha
    • one year ago
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    yes right so c^2=2 now take square root both sides bec you need c not c^2

  47. anonymous
    • one year ago
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    root2

  48. Nnesha
    • one year ago
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    yep right |dw:1438302263294:dw| so what is sin theta = ?

  49. Nnesha
    • one year ago
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    \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] remember this

  50. anonymous
    • one year ago
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    so sin theta = -1/root 2

  51. Nnesha
    • one year ago
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    yes right you can't leave the radical at the denominator so multiply both denoimnator and numerator by sqr root 2

  52. anonymous
    • one year ago
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    root 2 sin theta = 1?

  53. anonymous
    • one year ago
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    -1/2

  54. Nnesha
    • one year ago
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    correction:\[\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }\]

  55. anonymous
    • one year ago
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    so thats answer to cos?

  56. Nnesha
    • one year ago
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    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
    • one year ago
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    shoot this is hard

  58. anonymous
    • one year ago
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    idk i gtg

  59. anonymous
    • one year ago
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    thx tho

  60. Nnesha
    • one year ago
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    \[\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }\] multiply

  61. Nnesha
    • one year ago
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    alright i have to go,too we will continue later

  62. anonymous
    • one year ago
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    ok

  63. Nnesha
    • one year ago
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    :\[\frac{ -1 }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} }\] 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
    • one year ago
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    same way to find value of cos that's it!

  65. Nnesha
    • one year ago
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    if you have any question let me know tomorrow. :=)

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