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Mindblast3r

  • one year ago

help!

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  1. mindblast3r
    • one year ago
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    |dw:1436166754136:dw|

  2. Compassionate
    • one year ago
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    What do you need help solving

  3. mindblast3r
    • one year ago
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    find cosx

  4. Compassionate
    • one year ago
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    Do you know the relationship between cosx and sinx?

  5. mindblast3r
    • one year ago
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    no

  6. LynFran
    • one year ago
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    |dw:1436174086784:dw|

  7. imqwerty
    • one year ago
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    cosx = (root8)/3

  8. Compassionate
    • one year ago
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    The relationship between the cosine and sine graphs is that the cosine is the same as the sine — only it’s shifted to the left by 90 degrees, or π/2

  9. mindblast3r
    • one year ago
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    |dw:1436167048145:dw|

  10. mindblast3r
    • one year ago
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    x=2?

  11. LynFran
    • one year ago
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    |dw:1436174284089:dw|

  12. mindblast3r
    • one year ago
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    i don't even understand what i'm asking to be honest but i have a test on this soon. so pls explain to me what i'm asking.

  13. mindblast3r
    • one year ago
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    when i say,

  14. mindblast3r
    • one year ago
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    |dw:1436167321407:dw|

  15. mindblast3r
    • one year ago
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    what does this mean?

  16. mindblast3r
    • one year ago
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    sin x =1/3 Find cos x

  17. mindblast3r
    • one year ago
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    the question is that.

  18. mindblast3r
    • one year ago
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    i don't get what it's asking me??

  19. imqwerty
    • one year ago
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    \[cosx = 2\sqrt{2}/3\]

  20. Astrophysics
    • one year ago
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    Sinx = 1/3 that means the ratio opp/hyp -1/3, so we can make a triangle as following|dw:1436167626215:dw| now we need to figure out cosx, and remember the ratio of cosx is adj/hyp, so solve for the missing length using pythagorean theorem, and you will get your ratio, that make sense?

  21. mindblast3r
    • one year ago
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    DOES X MEAN INVERSE?

  22. Astrophysics
    • one year ago
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    Huh? No they put x there to represent it's an unkown, and that's what you're solving for

  23. Astrophysics
    • one year ago
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    |dw:1436167875535:dw| x is the adjacent side that will help you find cosx

  24. mindblast3r
    • one year ago
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    ohh it shows me the answer is |dw:1436167868096:dw|

  25. Astrophysics
    • one year ago
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    Well, lets see if its right

  26. Astrophysics
    • one year ago
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    We'll use \[a^2+b^2 = c^2\] where c is the hypotenuse. We'll let a = 1, b the side we are solving for, and c = 3 \[1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}\]

  27. Astrophysics
    • one year ago
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    So that's our x on the triangle |dw:1436168040178:dw| now we're looking for cosx right, \[cosx = \frac{ adj }{ hyp } = \frac{ \sqrt{8} }{ 3 }\]

  28. Astrophysics
    • one year ago
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    So yes, it simplifies to \[\frac{ 2\sqrt{2} }{ 3 }\]

  29. mindblast3r
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Astrophysics We'll use \[a^2+b^2 = c^2\] where c is the hypotenuse. We'll let a = 1, b the side we are solving for, and c = 3 \[1^2+b^2 = 3^2 \implies b^2= 3^2-1^2 \implies b=\sqrt{3^2-1^2} = \sqrt{9-1} = \sqrt{8}\] \(\color{blue}{\text{End of Quote}}\)

  30. Astrophysics
    • one year ago
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    Something not make sense there?

  31. mindblast3r
    • one year ago
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    wait i'm trying to understand.

  32. Astrophysics
    • one year ago
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    No worries, take your time

  33. anonymous
    • one year ago
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    Yeah, the pythagorean identity in terms of trig functions is written: \[ \sin^2x+\cos^2x =1 \]

  34. mindblast3r
    • one year ago
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    omg so many people viewing my question.

  35. UsukiDoll
    • one year ago
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    @imqwerty please refrain from giving direct answers because it's against OpenStudy's Code of Conduct. "Give Help, Not Answers I will encourage and guide those needing help, and not just give them an answer" "Don’t post only answers - guide the asker to a solution"

  36. Astrophysics
    • one year ago
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    Haha, don't get overwhelmed @Mindblast3r just see if you understand everything, if not ask questions

  37. UsukiDoll
    • one year ago
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    @Mindblast3r I know how you feel. I had 7 people watching me when I was tutoring sometimes. It was nervewrecking.

  38. anonymous
    • one year ago
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    You know, you can really only find \(|\cos x|\). You need more info to know the sign.

  39. mindblast3r
    • one year ago
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    |dw:1436168642801:dw|

  40. mindblast3r
    • one year ago
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    come*

  41. UsukiDoll
    • one year ago
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    that is from the hypotenuse

  42. UsukiDoll
    • one year ago
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    hypotenuse is the longest side of the triangle which was 3 cos x = adjacent / hypotenuse

  43. anonymous
    • one year ago
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    \[ |\cos x| = \sqrt{1 - \sin^2 x} \]

  44. UsukiDoll
    • one year ago
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    so when you were using the Pythagorean theorem, you were solving for one of the sides of the triangle.

  45. Astrophysics
    • one year ago
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    |dw:1436168854387:dw|

  46. imqwerty
    • one year ago
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    @UsukiDoll I am sorry :(

  47. Astrophysics
    • one year ago
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    What usuki said :), we were just solving for the adjacent side, I should've specified, then we used our cosx ratio adj/hyp

  48. Astrophysics
    • one year ago
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    where hyp = 3 and adj = sqrt(8)

  49. mindblast3r
    • one year ago
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    |dw:1436168916911:dw|

  50. UsukiDoll
    • one year ago
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    @wio 's method is another way of solving this problem, but you should only use it if you know the trig identities...otherwise it's best to use the Pythagorean theorem and drawing a triangle

  51. mindblast3r
    • one year ago
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    |dw:1436168957305:dw|

  52. Astrophysics
    • one year ago
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    Haha

  53. imqwerty
    • one year ago
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    ( ͡° ͜ʖ ͡°) u need cosx which is adjacent side/hypotenuse u knw that the hypotenuse is 3 nd nw u need adj. side u solve for adj. side by pythagorus theo. and plug it in the equation nd get answer=(root8)/3

  54. UsukiDoll
    • one year ago
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    @imqwerty that's ok...just remember next time that the main reason why OpenStudy exists is to provide guidance towards the question, not a massive online forum to trade answers.

  55. UsukiDoll
    • one year ago
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    my spelling and grammar is on vacation xD oy!

  56. anonymous
    • one year ago
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    The formula \[ \sin^2x+\cos^2x=1 \]Is based on the Pythagorean theorem.

  57. Astrophysics
    • one year ago
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    He's not there yet wio lol

  58. UsukiDoll
    • one year ago
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    If OP doesn't know those identities or if the problem specifies to use the Pythagorean theorem and draw the triangle method only, so be it.

  59. mindblast3r
    • one year ago
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    |dw:1436168987012:dw|

  60. Astrophysics
    • one year ago
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    Yes

  61. UsukiDoll
    • one year ago
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    yes @Mindblast3r :D

  62. imqwerty
    • one year ago
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    so very true

  63. mindblast3r
    • one year ago
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    |dw:1436169101210:dw|

  64. imqwerty
    • one year ago
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    problem is over nw party time ( ͡^ ͜ʖ ͡^)

  65. anonymous
    • one year ago
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    \[ (\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2 \\ \iff \frac{(\text{adjacent})^2}{(\text{hypotenuse})^2} + \frac{ (\text{opposite})^2 }{(\text{hypotenuse})^2} = 1 \\ \iff \left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)^2 + \left(\frac{\text{opposite}}{\text{hypotenuse}}\right)^2 = 1 \\ \iff \cos^2x+\sin^2x = 1 \]

  66. mindblast3r
    • one year ago
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    |dw:1436169251275:dw|

  67. Astrophysics
    • one year ago
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    Lol, I think usuki, Lyn, and qwerty should get a big thanks to :)

  68. mindblast3r
    • one year ago
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    |dw:1436169321629:dw|

  69. imqwerty
    • one year ago
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    ( ͡° ͜ʖ ͡°)

  70. imqwerty
    • one year ago
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    i met a man who said - "i have two daughters and both are girls."

  71. mindblast3r
    • one year ago
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    |dw:1436169398594:dw|

  72. mindblast3r
    • one year ago
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    ok people shows over, do something else.

  73. Astrophysics
    • one year ago
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    Haha ok, bye :). And a way I go... |dw:1436169509148:dw|

  74. mindblast3r
    • one year ago
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    lOL!

  75. mindblast3r
    • one year ago
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    Now that's funny!

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