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kaylala

  • 10 months ago

TOPIC: Trigonometry GIVEN THE FOLLOWING CONDITIONS: sin A = 4/5, A Σ I sin B = -3/5, B Σ III Find the value of: I. sin (A +- B) II. cos (A +- B) III. tan (A +- B)

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  1. kaylala
    • 10 months ago
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    @Callisto

  2. kaylala
    • 10 months ago
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    @eliassaab

  3. eliassaab
    • 10 months ago
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    I do not understand what comes after 4/5 and -3/5

  4. kaylala
    • 10 months ago
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    let me fix the question

  5. eliassaab
    • 10 months ago
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    Also look at this summary of trig formulas, your problem follows easily http://www.math.missouri.edu/escgi/mucgi-bin/trigidentities.cgi

  6. eliassaab
    • 10 months ago
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    It is still not fixed

  7. kaylala
    • 10 months ago
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    it is. see this pic that i will post. it's where i got the question. wait

  8. eliassaab
    • 10 months ago
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    You need these formulae sin(a + b ) = sin(a) cos(b) + sin(b) cos(a) sin(a - b ) = sin(a) cos(b) - sin(b) cos(a) cos(a + b ) = cos(a) cos(b) - sin(a) sin(b) cos(a - b ) = cos(a) cos(b) + sin(a) sin(b)

  9. kaylala
    • 10 months ago
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    oh it's (+_) the plus on top of the minus for I - III

  10. eliassaab
    • 10 months ago
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    You do first + than -

  11. eliassaab
    • 10 months ago
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    You need six answers

  12. eliassaab
    • 10 months ago
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    Where is the picture?

  13. kaylala
    • 10 months ago
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    wait. i just took a picture of it

  14. kaylala
    • 10 months ago
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    1 Attachment
  15. Yttrium
    • 10 months ago
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    @eliassaab , I think the information beside 4/5 and -3/5 are the position of angle in the cartesian plane. (specifically Quadrant I for <A and Quadrant III for <B

  16. kaylala
    • 10 months ago
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    @eliassaab here's the picture

  17. kaylala
    • 10 months ago
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    and yes those are values for the carthesian plane @Yttrium

  18. kaylala
    • 10 months ago
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    i dont think i have to find 6 answers @eliassaab

  19. kaylala
    • 10 months ago
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    somebody. anyone. please help

  20. Yttrium
    • 10 months ago
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    You really have to find 6 ansers @kaylala. Let me tell you this: \[\sin (A \pm B)\] is the notation for\[\sin (A+B)\] and \[\sin(A-B)\] I hope you understand now.

  21. kaylala
    • 10 months ago
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    oh okay i see now @Yttrium

  22. kaylala
    • 10 months ago
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    i just dont know how to start this

  23. Yttrium
    • 10 months ago
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    let's start with angle A

  24. Yttrium
    • 10 months ago
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    the condition says sinA = 4/5, where it is in quadrant 1. we know that sinx = opp/hyp Hence, the figure much look like this. |dw:1391508636187:dw|

  25. kaylala
    • 10 months ago
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    okay

  26. Yttrium
    • 10 months ago
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    |dw:1391508787128:dw| let's say that adjacent side of A is x. Can you solve x now?

  27. kaylala
    • 10 months ago
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    okay x=sqrt-9 ??? @Yttrium

  28. Yttrium
    • 10 months ago
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    isn't is sqrt(+9)? which is 3?

  29. kaylala
    • 10 months ago
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    oh my bad

  30. kaylala
    • 10 months ago
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    it do is 3

  31. Yttrium
    • 10 months ago
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    Yea it is. And, with that you can now actually solve for cosA, tanA.

  32. kaylala
    • 10 months ago
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    how?

  33. kaylala
    • 10 months ago
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    i really dont know how to do this please guide me @Yttrium

  34. Yttrium
    • 10 months ago
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    cosA = adj/hyp tanA = opp/adj did you forget that?

  35. kaylala
    • 10 months ago
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    sorta

  36. kaylala
    • 10 months ago
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    cos A = 3/5 ???

  37. kaylala
    • 10 months ago
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    tanA=4/3

  38. kaylala
    • 10 months ago
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    did i get it?

  39. kaylala
    • 10 months ago
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    what do i do next?

  40. Yttrium
    • 10 months ago
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    You got this. I think it is better if you write sinA, cosA, and tanA in a piece of paper. They are really important later. Next thing we are going to do is to solve for cosB and tanB

  41. kaylala
    • 10 months ago
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    yeah. just wrote it down already how do i solve for that?

  42. kaylala
    • 10 months ago
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    y=4

  43. kaylala
    • 10 months ago
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    ? @Yttrium

  44. Yttrium
    • 10 months ago
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    you sure? isn't it -4? take note it is in the quadrant III.

  45. kaylala
    • 10 months ago
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    okay so the quadrant determines the sign? right?

  46. Yttrium
    • 10 months ago
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    Yeah. for (x,y) quadrant I (+,+) II (-,+) III (-,-) IV (+,-) Now. Find cosB and tanB. Be careful with the signs.

  47. kaylala
    • 10 months ago
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    okay then

  48. Yttrium
    • 10 months ago
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    Solve for cosB and tanB

  49. kaylala
    • 10 months ago
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    cos B= -4/5

  50. kaylala
    • 10 months ago
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    tanB= 3/-4

  51. Yttrium
    • 10 months ago
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    Ohhh wait!!!

  52. kaylala
    • 10 months ago
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    did i get it?

  53. kaylala
    • 10 months ago
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    ookay

  54. Yttrium
    • 10 months ago
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    the graph will be like this |dw:1391510194742:dw|

  55. Yttrium
    • 10 months ago
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    x then is -4 solve again for cosB and tanB

  56. kaylala
    • 10 months ago
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    i see now but the answer for cos b and tan B still remains?

  57. kaylala
    • 10 months ago
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    so are those the answers already?

  58. kaylala
    • 10 months ago
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    what to do next?

  59. kaylala
    • 10 months ago
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    @Yttrium

  60. Yttrium
    • 10 months ago
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    ooops tanB = -3/-4 = 3/4

  61. kaylala
    • 10 months ago
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    oh you're right

  62. Yttrium
    • 10 months ago
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    Now is the time to evaluate them. I. sin (A +- B) = sinAcosB +- cosAsinB II. cos (A +- B) = cosAcos -+ sinAsinB

  63. Yttrium
    • 10 months ago
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    III. \[\tan(A \pm B) = \frac{ tanA \pm tanB }{ 1\mp tanAtanB }\]

  64. kaylala
    • 10 months ago
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    how's that? do i just substitute what i got? to the given formulas?

  65. Yttrium
    • 10 months ago
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    Yeah. All the operation found upside are the only operations used when applying upside identities. Look at this. \[\tan(A +B) = \frac{ tanA + tanB }{ 1 - tanAtanB }\]

  66. kaylala
    • 10 months ago
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    so sin(a+b) = 1 sin(a-b) = -7/25 ??? @Yttrium

  67. kaylala
    • 10 months ago
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    cos (a+b)=0 cos (a-b)=-24/25 did i get it correctly @Yttrium ?

  68. Yttrium
    • 10 months ago
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    i'll just check the sin function, you take care of the others :D sin(a+b) = -1

  69. kaylala
    • 10 months ago
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    okay :) thanks

  70. kaylala
    • 10 months ago
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    tan (A+B)= undefined??? is it? please check this @Yttrium

  71. Yttrium
    • 10 months ago
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    what is tanA and tanbB?

  72. kaylala
    • 10 months ago
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    tanA=4/3 tanB=3/4

  73. Yttrium
    • 10 months ago
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    Yeah. That's undefined. And that is normal. :))

  74. kaylala
    • 10 months ago
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    i see. okay thanks again :)

  75. Yttrium
    • 10 months ago
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    No problem :))

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