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kaylala Group Title

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)

  • 7 months ago
  • 7 months ago

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  1. kaylala Group Title
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    @Callisto

    • 7 months ago
  2. kaylala Group Title
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    @eliassaab

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

    • 7 months ago
  4. kaylala Group Title
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    let me fix the question

    • 7 months ago
  5. eliassaab Group Title
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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

    • 7 months ago
  6. eliassaab Group Title
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    It is still not fixed

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

    • 7 months ago
  8. eliassaab Group Title
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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)

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

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

    • 7 months ago
  11. eliassaab Group Title
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    You need six answers

    • 7 months ago
  12. eliassaab Group Title
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    Where is the picture?

    • 7 months ago
  13. kaylala Group Title
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    wait. i just took a picture of it

    • 7 months ago
  14. kaylala Group Title
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    • 7 months ago
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  15. Yttrium Group Title
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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

    • 7 months ago
  16. kaylala Group Title
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    @eliassaab here's the picture

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

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

    • 7 months ago
  19. kaylala Group Title
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    somebody. anyone. please help

    • 7 months ago
  20. Yttrium Group Title
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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.

    • 7 months ago
  21. kaylala Group Title
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    oh okay i see now @Yttrium

    • 7 months ago
  22. kaylala Group Title
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    i just dont know how to start this

    • 7 months ago
  23. Yttrium Group Title
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    let's start with angle A

    • 7 months ago
  24. Yttrium Group Title
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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|

    • 7 months ago
  25. kaylala Group Title
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    okay

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

    • 7 months ago
  27. kaylala Group Title
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    okay x=sqrt-9 ??? @Yttrium

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

    • 7 months ago
  29. kaylala Group Title
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    oh my bad

    • 7 months ago
  30. kaylala Group Title
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    it do is 3

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

    • 7 months ago
  32. kaylala Group Title
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    how?

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

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

    • 7 months ago
  35. kaylala Group Title
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    sorta

    • 7 months ago
  36. kaylala Group Title
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    cos A = 3/5 ???

    • 7 months ago
  37. kaylala Group Title
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    tanA=4/3

    • 7 months ago
  38. kaylala Group Title
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    did i get it?

    • 7 months ago
  39. kaylala Group Title
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    what do i do next?

    • 7 months ago
  40. Yttrium Group Title
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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

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

    • 7 months ago
  42. kaylala Group Title
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    y=4

    • 7 months ago
  43. kaylala Group Title
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    ? @Yttrium

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

    • 7 months ago
  45. kaylala Group Title
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    okay so the quadrant determines the sign? right?

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

    • 7 months ago
  47. kaylala Group Title
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    okay then

    • 7 months ago
  48. Yttrium Group Title
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    Solve for cosB and tanB

    • 7 months ago
  49. kaylala Group Title
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    cos B= -4/5

    • 7 months ago
  50. kaylala Group Title
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    tanB= 3/-4

    • 7 months ago
  51. Yttrium Group Title
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    Ohhh wait!!!

    • 7 months ago
  52. kaylala Group Title
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    did i get it?

    • 7 months ago
  53. kaylala Group Title
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    ookay

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

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

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

    • 7 months ago
  57. kaylala Group Title
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    so are those the answers already?

    • 7 months ago
  58. kaylala Group Title
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    what to do next?

    • 7 months ago
  59. kaylala Group Title
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    @Yttrium

    • 7 months ago
  60. Yttrium Group Title
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    ooops tanB = -3/-4 = 3/4

    • 7 months ago
  61. kaylala Group Title
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    oh you're right

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

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

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

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

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

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

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

    • 7 months ago
  69. kaylala Group Title
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    okay :) thanks

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

    • 7 months ago
  71. Yttrium Group Title
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    what is tanA and tanbB?

    • 7 months ago
  72. kaylala Group Title
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    tanA=4/3 tanB=3/4

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

    • 7 months ago
  74. kaylala Group Title
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    i see. okay thanks again :)

    • 7 months ago
  75. Yttrium Group Title
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    No problem :))

    • 7 months ago
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