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lilsis76

  • 3 years ago

Given tan theta = -2/3, where pi/2 < 0 < pi and csc B=2 where 0<B<pi/2 Find sin (0+B) and cos (B-0) ill draw the 3 data thingys given

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  1. lilsis76
    • 3 years ago
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    |dw:1350692195063:dw|

  2. surdawi
    • 3 years ago
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    |dw:1350692395385:dw|

  3. lilsis76
    • 3 years ago
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    based on the information i know it will be first quad.

  4. surdawi
    • 3 years ago
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    csc B = 1/sin B

  5. lilsis76
    • 3 years ago
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    oh duh haha its negative, so its going backwards right?

  6. surdawi
    • 3 years ago
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    yes

  7. lilsis76
    • 3 years ago
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    |dw:1350692578128:dw| right?

  8. surdawi
    • 3 years ago
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    yes, next

  9. lilsis76
    • 3 years ago
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    dang it, im going to need identities right? cuz it kinda looks like it :/

  10. lilsis76
    • 3 years ago
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    im trying to find the identities

  11. surdawi
    • 3 years ago
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    1st find theta and beta sin (0 + B) = sin 0 cos B + cos 0 sin B cos (0 – B) = cos 0 cos B + sin 0 sin B

  12. lilsis76
    • 3 years ago
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    UGH!!! okay im back lol okay, are these the identities I should know for finding theta and beta then? just so i save it?

  13. surdawi
    • 3 years ago
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    \[\tan \theta=\frac{ -2 }{ 3 }\] \[\tan^{-1} (\tan \theta)=\tan^{-1} (\frac{ -2 }{ 3 })\]

  14. lilsis76
    • 3 years ago
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    okay so far i know: sin = 2/x cos =2/x and tan is the same -2/3

  15. surdawi
    • 3 years ago
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    what does x = ?

  16. surdawi
    • 3 years ago
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    |dw:1350693950387:dw|

  17. surdawi
    • 3 years ago
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    by the way what grade is this

  18. lilsis76
    • 3 years ago
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    okay hold on, this is precal. im in a community college, but im horrible at math haha

  19. lilsis76
    • 3 years ago
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    x = sqrt 13

  20. surdawi
    • 3 years ago
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    been a while since i been in the class anyways x is correct

  21. lilsis76
    • 3 years ago
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    oh really? did u graduate? and YAY okay i got the x, now will that be the theta?

  22. lilsis76
    • 3 years ago
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    okay so the cos=3sqrt13 /26

  23. surdawi
    • 3 years ago
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    right now im working on my|dw:1350694287644:dw|

  24. surdawi
    • 3 years ago
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    can you calculate theta now

  25. lilsis76
    • 3 years ago
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    uh.... well the one i had or placed on here was wrong? the okay so the cos=3sqrt13 /26

  26. lilsis76
    • 3 years ago
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    is the 2 a regular 2, or a negative 2? since its going downward of the unit circle?

  27. lilsis76
    • 3 years ago
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    wait im lost: i want to go back a lil to the soh cah toa cah is adj/hyp that would be 3 / sqrt13 |dw:1350694753458:dw| right?

  28. surdawi
    • 3 years ago
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    thats the cos of thata, dont you want theta?

  29. lilsis76
    • 3 years ago
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    yes the theta of tangent

  30. lilsis76
    • 3 years ago
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    wait. cos theta = the 3sqrt13 / 13

  31. lilsis76
    • 3 years ago
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    im so lost

  32. surdawi
    • 3 years ago
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    can i do it on a piece of paper then send it to you, im tired and want to go to sleep; yes i know its only 9:20 lol

  33. lilsis76
    • 3 years ago
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    lol yes you may. and thanks, im getting a headache from this too

  34. Outkast3r09
    • 3 years ago
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    alright so first we need to figure out where tan is negative. Using this circle |dw:1350696636481:dw|

  35. Outkast3r09
    • 3 years ago
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    these are when the functions are positive. for instance, in quadrant 1 , all function are positive

  36. lilsis76
    • 3 years ago
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    okay so tan would be negative in...................the cosine area

  37. Outkast3r09
    • 3 years ago
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    and also the sine , because tangent is \[\frac{sin\theta}{cos\theta}\]... so anytime either one is negative tan will be negative ... when sin and cos are both negative ,tan is positive which is this quadrant|dw:1350696881960:dw|

  38. lilsis76
    • 3 years ago
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    okay so the 2nd and 3rd piece on the left sides

  39. Outkast3r09
    • 3 years ago
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    next we use the information it gives about theta \[\pi/2<\theta<\pi\] can you mark these angles on my picture

  40. lilsis76
    • 3 years ago
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    |dw:1350697001280:dw|

  41. Outkast3r09
    • 3 years ago
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    so theta is less than \[\pi/2\] but greater than \[\pi\]|dw:1350697010218:dw|

  42. Outkast3r09
    • 3 years ago
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    this means \[\theta is in your second quadrant

  43. lilsis76
    • 3 years ago
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    okay i see that now

  44. Outkast3r09
    • 3 years ago
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    alright so now lets solve the equation \[tan(\theta)=\frac{-2}{3}\]

  45. Outkast3r09
    • 3 years ago
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    how would you get theta by itself

  46. Outkast3r09
    • 3 years ago
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    do you know about inverse trig functions?

  47. lilsis76
    • 3 years ago
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    im back server was werid. let me look

  48. lilsis76
    • 3 years ago
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    no i dont know about the inverse trig functions unless your talkin about it being 1/tan or like the cos/sin

  49. Outkast3r09
    • 3 years ago
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    inverse function work as "cancelers" of trig functions they're usually denoted as \[arcsin,arctan,arccos\] or \[sin^{-1},cos^{-1}\] you know them because you know 1/tanx. I think you just didn't know what they were called

  50. Outkast3r09
    • 3 years ago
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    these cancel out the same trig function so \[arcsin(sin(x))=x\]

  51. Outkast3r09
    • 3 years ago
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    so what would you do in your case \[tan(\theta)=\frac{-2}{3}\]

  52. lilsis76
    • 3 years ago
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    no we have not gone over this :/ and so the arctan one would be arctan(tan(x))= -2/3 ?

  53. Outkast3r09
    • 3 years ago
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    if you do arctan on one side you have to do it on the other =]

  54. lilsis76
    • 3 years ago
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    -2/3arctan(tan(x))=-2/3

  55. Outkast3r09
    • 3 years ago
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    ok say \[tan(x)=\frac{-2}{3}\] you know to get rid of tan you have to use arctan, however if you multiply the left side by arctan, you have to multiply the right side also \[arctan(tan(x)=arctan(\frac{-2}{3})\]

  56. Outkast3r09
    • 3 years ago
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    since arctan cancels tan you get \[x=arctan(\frac{-2}{3})\]

  57. Outkast3r09
    • 3 years ago
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    your calculator should have an arctan symbol. it might be \[tan^{-1}\]

  58. lilsis76
    • 3 years ago
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    so the tan^-1 is called arctan?

  59. Outkast3r09
    • 3 years ago
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    yes =]

  60. lilsis76
    • 3 years ago
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    weird....okay, ya i didnt know any inverse ones i couldnt find the identities either :/ but i did tan^-1(-2/3) and it gave me.............-33.69

  61. Outkast3r09
    • 3 years ago
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    alright where is -33.69 degrees?

  62. lilsis76
    • 3 years ago
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    haha you really ask me degreees?? umm... i think its that 180/pi, right?

  63. Outkast3r09
    • 3 years ago
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    theta is in degrees unless you put in radians but keep it in degrees , all i'm asking is where on a circle would -33.69 degrees be( quadrant wise)

  64. lilsis76
    • 3 years ago
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    OH, umm... well hold on

  65. lilsis76
    • 3 years ago
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    |dw:1350698454198:dw|

  66. Outkast3r09
    • 3 years ago
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    well uh oh that isn't in the second quadrant we need to find an angle that is in the second quadrant with the same value as -33.69 degrees

  67. lilsis76
    • 3 years ago
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    should i start from 90 degrees?? |dw:1350698677314:dw|

  68. Outkast3r09
    • 3 years ago
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    hold on i know what you need to do just don't know why it's been a while since going through this material

  69. Outkast3r09
    • 3 years ago
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    i know you have to add pi(180 degrees) to it

  70. lilsis76
    • 3 years ago
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    okay, i think its mult. the ( 180/pi) i think

  71. lilsis76
    • 3 years ago
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    i cant find the equ.

  72. lilsis76
    • 3 years ago
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    WAIT!!! to change to degrees you multiply 180 deg/ pi. but i would need the point for the other one to do it..... here is an example from my notes: pi/12(180/pi) cross reduce to--->180 deg./12 = 15 degrees---sin15degrees =.2588

  73. lilsis76
    • 3 years ago
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    I really cant figure this one out... can you? or are you writingit down on paper, or looking it up online??

  74. jim_thompson5910
    • 3 years ago
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    in your drawing at the very top, you say cos 7/25 did you mean cos(theta) = 7/25 ???

  75. lilsis76
    • 3 years ago
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    yes

  76. jim_thompson5910
    • 3 years ago
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    so because sin^2(x) + cos^2(x) = 1 you can find sin(theta) like so sin^2(theta) + cos^2(theta) = 1 sin^2(theta) + (7/25)^2 = 1 sin^2(theta) + 49/625 = 1 sin^2(theta) = 1 - 49/625 sin^2(theta) = 576/625 sin(theta) = -sqrt(576/625) sin(theta) = -24/25

  77. jim_thompson5910
    • 3 years ago
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    oh wait, sry sine is positive because 0 < theta < pi/2

  78. jim_thompson5910
    • 3 years ago
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    so sin(theta) = 24/25

  79. lilsis76
    • 3 years ago
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    lol let me check

  80. jim_thompson5910
    • 3 years ago
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    sin^2(B) + cos^2(B) = 1 sin^2(B) + (-3/5)^2 = 1 sin^2(B) + 9/25 = 1 sin^2(B) = 1 - 9/25 sin^2(B) = 16/25 sin(B) = -sqrt(16/25) ... sin(B) is negative for pi < B < 3pi/2 sin(B) = -4/5

  81. jim_thompson5910
    • 3 years ago
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    So after those two steps, we now know these four pieces of info sin(theta) = 24/25 cos(theta) = 7/25 sin(B) = -4/5 cos(B) = -3/25 You can now use them in the identities sin(theta + B) = sin(theta)cos(B) + cos(theta)sin(B) cos(theta - B) = cos(theta)cos(B) + sin(theta)sin(B)

  82. lilsis76
    • 3 years ago
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    wait sin B is negative?? because..... its in the ..3rd quardinate and sin is negative and 24/25 is positive because it falls between the |dw:1350700800280:dw| 0 < theta < pi/2

  83. jim_thompson5910
    • 3 years ago
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    yeah i fixed that error, my bad

  84. lilsis76
    • 3 years ago
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    lol just a moment

  85. lilsis76
    • 3 years ago
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    wait sin B is negative?? because..... its in the ..3rd quardinate and sin is negative and 24/25 is positive because it falls between the |dw:1350700800280:dw| 0 < theta < pi/2

  86. lilsis76
    • 3 years ago
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    Sin B is negative??? because its in the 3rd quar. and that sin is negative and 24/25 is positive because it falls between the 0 < theta < pi/2

  87. jim_thompson5910
    • 3 years ago
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    yeah i fixed that

  88. lilsis76
    • 3 years ago
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    okay, im checking my notes cuz i have writing in every direction

  89. lilsis76
    • 3 years ago
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    okay, thank you, lol i have to write it neater and go over it again to get things down. thank you for helping me. ill give u a medal

  90. jim_thompson5910
    • 3 years ago
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    yw

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