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anonymous

  • 5 years ago

Use an addition or subtraction formula to find the exact value of the expression. tan(-5pi/12) sin(23/12) cos(13pi/12)

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  1. anonymous
    • 5 years ago
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    Please help!

  2. amistre64
    • 5 years ago
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    it wants you to seperate the "radian" given into known radians that you can play with

  3. amistre64
    • 5 years ago
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    like pi/3 pi/4 pi/6 pi/ 2pi those things

  4. amistre64
    • 5 years ago
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    3 and 2 = 5..... lets try that on the first one

  5. anonymous
    • 5 years ago
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    OK

  6. amistre64
    • 5 years ago
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    a=-3pi/12 b=2pi/12 a = -pi/4 b = pi/6

  7. anonymous
    • 5 years ago
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    still remember the addition stuff for tangent?? =D

  8. anonymous
    • 5 years ago
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    Kind of

  9. amistre64
    • 5 years ago
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    tana + tanb --------- ;) 1 - tanatanb

  10. amistre64
    • 5 years ago
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    but we is subtracting here.... or adding if you wanna get too technicalized about it

  11. amistre64
    • 5 years ago
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    lets add....i wanna add :)

  12. amistre64
    • 5 years ago
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    tan(-pi/4) + tan(-pi/6) --------------------- 1 - tan(-pi/4) tan(-pi/6)

  13. anonymous
    • 5 years ago
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    lets add :)

  14. anonymous
    • 5 years ago
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    where did u get all of these numbers from

  15. amistre64
    • 5 years ago
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    -1 + -sqrt(3)/3 ----------------- 1 - (-1)(-sqrt(3)/3)

  16. anonymous
    • 5 years ago
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    Huh?

  17. amistre64
    • 5 years ago
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    -5pi -3pi -2pi ---- = ---- + ---- right? 12 12 12

  18. anonymous
    • 5 years ago
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    yes

  19. amistre64
    • 5 years ago
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    goos, the reduce those 2 new radians to something we can use: something familiar.. -pi/4 and -pi/6

  20. amistre64
    • 5 years ago
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    tan(-pi/4) = -1 its a 45 degree line with a negative slope tan(-pi/6) = -30degrees = -1/sqrt(3) = sqrt(3)/3

  21. amistre64
    • 5 years ago
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    -1 + -sqrt(3)/3 --------------- 1 - (-1)(-sqrt(3)/3)

  22. anonymous
    • 5 years ago
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    ok

  23. amistre64
    • 5 years ago
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    -1 - sqrt(3)/3 -3 - sqrt(3) ------------ = ----------- we can rationalize the 1 - sqrt(3)/3 3 - sqrt(3) denominator maybe :)

  24. amistre64
    • 5 years ago
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    -3 - sqrt(3) (3+sqrt(3)) -12 -6sqrt(3) ----------- --------- = ------------- 3 - sqrt(3) (3+sqrt(3)) 6 reducing we get: -2 -sqrt(3) if I did it right in me head

  25. anonymous
    • 5 years ago
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    yes

  26. amistre64
    • 5 years ago
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    yes im right? or yes if we agree theyll shut up? :)

  27. anonymous
    • 5 years ago
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    um... both

  28. amistre64
    • 5 years ago
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    lol .... i got enough voices in me head that we aint gonna shut up regardless lol

  29. anonymous
    • 5 years ago
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    lol

  30. amistre64
    • 5 years ago
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    sin(23/12) is this missing a pi?

  31. anonymous
    • 5 years ago
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    ok so, the answer for the first one is -2sqrt3

  32. amistre64
    • 5 years ago
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    no its: (-2) - (sqrt(3))

  33. anonymous
    • 5 years ago
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    Yes, it is. It is sin(23pi/12)

  34. amistre64
    • 5 years ago
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    23 is almost 24; so were looking at 2pi - pi/12 at that one. but pi/12 aint in our "easy" angles so we gotta look harder for an answer

  35. anonymous
    • 5 years ago
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    ok

  36. anonymous
    • 5 years ago
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    why r u rounding up to 24?

  37. amistre64
    • 5 years ago
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    just looking at where the angle ends up; its tucked under the 2pi. 23pi/12 = 20pi/12 + 3pi/12

  38. amistre64
    • 5 years ago
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    20pi/12 = 5pi/3 3pi/12 = pi/4

  39. amistre64
    • 5 years ago
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    these are angles we know about, so we can use these

  40. anonymous
    • 5 years ago
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    ok

  41. amistre64
    • 5 years ago
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    sin(23pi/12) = sin(5pi/3 + pi/4) sin(a+b) = sin(a)cos(b)+sin(b)cos(a)

  42. amistre64
    • 5 years ago
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    sin(5pi/3) = -sqrt(3)/2 cos(5pi/3) = 1/2 sin(pi/4) = cos(pi/4) = 1/sqrt(2)

  43. anonymous
    • 5 years ago
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    is the answer [sqrt3-1]/ [2sqrt2]

  44. amistre64
    • 5 years ago
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    sin(a+b) = (-sqrt(2)/2) (1/sqrt(2)) +(1/sqrt(2))(1/2)

  45. amistre64
    • 5 years ago
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    -2/4 + sqrt(2)/4 = sqrt(2)-2 -------- 4

  46. amistre64
    • 5 years ago
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    ack!!...i typoed that up didnt I

  47. anonymous
    • 5 years ago
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    yes

  48. amistre64
    • 5 years ago
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    (-sqrt(3)/2) (1/sqrt(2)) +(1/sqrt(2))(1/2) -sqrt(6)/4 + sqrt(2)/4 = sqrt(2) - sqrt(6) -------------- 4

  49. amistre64
    • 5 years ago
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    that should be correct :)

  50. anonymous
    • 5 years ago
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    let's see...

  51. amistre64
    • 5 years ago
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    im going for coffee......like a bad little mormon :)

  52. anonymous
    • 5 years ago
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    Correct!

  53. anonymous
    • 5 years ago
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    haha...

  54. anonymous
    • 5 years ago
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    have we done the third one?

  55. amistre64
    • 5 years ago
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    theres a third one? and the watering hole is getting refilled .....

  56. anonymous
    • 5 years ago
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    you* sorry for taking the credit :P

  57. anonymous
    • 5 years ago
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    It is cos(13pi/12)

  58. amistre64
    • 5 years ago
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    your like my conscinece....that is if I had a consicence lol

  59. amistre64
    • 5 years ago
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    pi/12 = 15 degrees. I knew it!!

  60. amistre64
    • 5 years ago
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    this is 225 - 30 degrees.... or 5pi/4 - pi/6

  61. amistre64
    • 5 years ago
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    cos(a-b) - cos(a)cos(b) + sin(a)sin(b)

  62. amistre64
    • 5 years ago
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    sin and cos 5pi/4 = -1/sqrt(2) (a) sin 30 = 1/2 cos30 = sqrt(3)/2

  63. anonymous
    • 5 years ago
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    How the hell do you know all of this?

  64. amistre64
    • 5 years ago
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    (-1/sqrt(2)) (sqrt3)/2) + (-1/sqrt(2)) (1/2) -sqrt(6)/4 + -sqrt(2)/4 = -sqrt(6) -sqrt(2) -------------- 4

  65. amistre64
    • 5 years ago
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    I know nutheeeng!!

  66. anonymous
    • 5 years ago
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    kidding me

  67. amistre64
    • 5 years ago
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    col. klink right?

  68. anonymous
    • 5 years ago
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    can u help me with more? PLEASEEEEEEE:)

  69. amistre64
    • 5 years ago
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    was I right?

  70. anonymous
    • 5 years ago
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    Oh yeah....

  71. amistre64
    • 5 years ago
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    how am i supoosed to increase my fan numbers if I only help one person?......lol, like that matters HAH!!; yeah, I can help with what I can;)

  72. anonymous
    • 5 years ago
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    THANK YOU SO MUCH!!!

  73. amistre64
    • 5 years ago
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    its all about the benjamins baybay lol

  74. anonymous
    • 5 years ago
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    Use an addition or subtraction formula to find the exact value of the expression. Tan(-15degrees)

  75. anonymous
    • 5 years ago
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    yep

  76. amistre64
    • 5 years ago
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    sin juice is sitting on the table again; brb....

  77. anonymous
    • 5 years ago
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    ok

  78. amistre64
    • 5 years ago
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    the tan(-15) degrees = tan(-pi/12) = tan(23pi/12)

  79. amistre64
    • 5 years ago
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    lets go with 45 - 30; cause 15 = pi/12

  80. anonymous
    • 5 years ago
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    Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value. cos (17pi/21) cos (-pi/7) - sin (17pi/21) sin (-pi/7) This is gibberish to me :'(

  81. anonymous
    • 5 years ago
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    ok

  82. amistre64
    • 5 years ago
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    im gonna guess that the tan(-15) = -sqrt(3)/6 but let me verify that with my sources ;)

  83. anonymous
    • 5 years ago
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    you're your own source lol

  84. anonymous
    • 5 years ago
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    it's wrong, my hw program says it's wrong

  85. amistre64
    • 5 years ago
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    I aint verified it yet....hold on lol

  86. amistre64
    • 5 years ago
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    tan(45) = 1 ; tan(30) = 1/sqrt(3) tan(45-30) = tan(45) - tan(30) --------------- 1 + tan(45)tan(30) 1 - sqrt(3)/3 (1-sqrt(3)/3) ----------- ----------- 1 + sqrt(3)/3 (1-sqrt(3)/3)

  87. anonymous
    • 5 years ago
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    ok

  88. amistre64
    • 5 years ago
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    1 -2sqrt(3)/3 + 3/9 ------------------ 1 - 3/9

  89. amistre64
    • 5 years ago
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    9 -6sqrt(3) + 3 ----------- 9 ------------- 9-3 --- 9

  90. amistre64
    • 5 years ago
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    9 -6sqrt(3) + 3 12 -6sqrt(3) ------------- = ----------- = 2 -sqrt(3) 9-3 6 but thats (15) degrees, we need to negate it -1(2 -sqrt(3)) = sqrt(3) - 2

  91. anonymous
    • 5 years ago
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    Righto... :)

  92. amistre64
    • 5 years ago
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    i read it in the stars lol

  93. anonymous
    • 5 years ago
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    Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value. cos (17pi/21) cos (-pi/7) - sin (17pi/21) sin (-pi/7) This is gibberish to me :'(

  94. anonymous
    • 5 years ago
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    haha...

  95. amistre64
    • 5 years ago
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    cos(a+b) = cos(a)cos(b) - sin(a)sin(b) they give you "a" and "b" so.... cos(17pi/21 + -pi/7) cos(17pi/21 -3pi/21) = cos(14pi/21) = cos(2pi/3) or cos(120) degrees

  96. amistre64
    • 5 years ago
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    cos(120) = -1/2

  97. anonymous
    • 5 years ago
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    how do you get cos120 = -1/2

  98. amistre64
    • 5 years ago
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    how? cause 120 degree is in Q2 were cos is negative. so just take the reference angle of cos(60) and negate it. cos(60) = 1/2; cos(120) = -1/2

  99. anonymous
    • 5 years ago
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    oh ok...

  100. amistre64
    • 5 years ago
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    weird als got a paradoy called "its all about the pentiums baby"! its hilarious

  101. anonymous
    • 5 years ago
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    haha... lol. Is he still around?

  102. anonymous
    • 5 years ago
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    This is hilarious http://www.youtube.com/watch?v=3u-J4uxEWFQ

  103. amistre64
    • 5 years ago
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    the hit i put out on him hasnt been paid yet, so yeah :)

  104. anonymous
    • 5 years ago
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    Damn it!

  105. amistre64
    • 5 years ago
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    libraries conputers to old....says it needs a java update but wont take it :/

  106. anonymous
    • 5 years ago
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    ah... what school or college do u go to?

  107. amistre64
    • 5 years ago
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    phcc down here in florida

  108. anonymous
    • 5 years ago
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    oh.. ok

  109. anonymous
    • 5 years ago
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    i'm at umd- college park

  110. amistre64
    • 5 years ago
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    maryland?

  111. anonymous
    • 5 years ago
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    yeah

  112. amistre64
    • 5 years ago
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    annapolis is your capital lol

  113. anonymous
    • 5 years ago
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    good job :) A+

  114. anonymous
    • 5 years ago
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    u should be at MIT or CAL-TECH lol

  115. amistre64
    • 5 years ago
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    it took me 39 years to memorize the states and their capitals :) now its just useless junk rattling around...

  116. anonymous
    • 5 years ago
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    yep...

  117. anonymous
    • 5 years ago
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    my parents have been telling me that since i was born lol this stuff will get you somewhere not remembering capitals

  118. anonymous
    • 5 years ago
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    but back to the problem...

  119. amistre64
    • 5 years ago
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    which problem we got now?

  120. anonymous
    • 5 years ago
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    Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value. cos (17pi/21) cos (-pi/7) - sin (17pi/21) sin (-pi/7) This is gibberish to me :'(

  121. amistre64
    • 5 years ago
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    dint we just finish that one?

  122. amistre64
    • 5 years ago
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    cos(120) = -1/2

  123. anonymous
    • 5 years ago
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    sorry we u did that one already

  124. anonymous
    • 5 years ago
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    Simplify the trigonometric expression. cos(x)/sec(x) + sin(x)/csc(x)

  125. amistre64
    • 5 years ago
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    its like im stuck in a time warp that leads thru the twilight zone .....

  126. amistre64
    • 5 years ago
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    is that a left term + a right term? or is that right side all under the fraction bar? use them () to seperate things up

  127. anonymous
    • 5 years ago
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    it is a left arm and a right arm

  128. anonymous
    • 5 years ago
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    \[\cos [x]\div \sec [x] +\sin [x]\div \csc [x]\] If this helps

  129. amistre64
    • 5 years ago
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    sec = 1/cos csc = 1/sin 1+1 = 2

  130. amistre64
    • 5 years ago
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    next!!

  131. anonymous
    • 5 years ago
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    Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value. tan(pi/3) + tan(5pi/6) / 1- tan(pi/3) tan(5pi/6)

  132. amistre64
    • 5 years ago
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    thats the expanded version of tan(a+b) a = pi/3 b = 5pi/6

  133. anonymous
    • 5 years ago
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    ok

  134. amistre64
    • 5 years ago
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    pi/3 + 5pi/6 = 2pi +... = 7pi/6 = pi + pi/6

  135. amistre64
    • 5 years ago
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    tan(7pi/6) = tan(pi/6) = tan(30) = 1/sqrt(3) = sqrt(3)/3

  136. anonymous
    • 5 years ago
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    1) Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value. cos(10) cos(80) - sin(10) sin(80) 2) Use an addition or subtraction formula to find the exact value of the expression. sin(15)

  137. amistre64
    • 5 years ago
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    1 is a cos(a+b) cos(10+80) = cos(90) = 0 2) sin(pi/12)..dint we do that one earlier?

  138. anonymous
    • 5 years ago
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    no

  139. amistre64
    • 5 years ago
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    it looks familiar :)

  140. anonymous
    • 5 years ago
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    Use an addition or subtraction formula to find the exact value of the expression. sin(15)

  141. amistre64
    • 5 years ago
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    we did the cos(13/12pi)...which is pretty much what we want with a sign adjustment

  142. amistre64
    • 5 years ago
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    sqrt(6)+sqrt(2) cos(pi/12) = -------------- 4 cos^2 + sin^2 = 1 sqrt(6)+sqrt(2) 8+4sqrt(3) -------------- = ----------- 4 16 8+4sqrt(3) 1 - ----------- = sin^2(15) 16

  143. amistre64
    • 5 years ago
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    8 -4sqrt(3) ------------- = sin^2(15) 16

  144. amistre64
    • 5 years ago
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    4 (2 -sqrt(3)) sqrt ------------- = sin(15) 16

  145. amistre64
    • 5 years ago
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    sqrt(2-sqrt(3)) ------------- = sin(15) maybe? 2

  146. anonymous
    • 5 years ago
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    that doesn't work

  147. amistre64
    • 5 years ago
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    \[\sqrt{2 - \sqrt{3}}/2\]

  148. amistre64
    • 5 years ago
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    how you get a bar for a fraction in this thing?

  149. anonymous
    • 5 years ago
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    u can't i tried doing it to

  150. amistre64
    • 5 years ago
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    if anything; sin(45-30) = sin(15) sin(45-30) = sin(1/sqrt(2))cos(1/sqrt(3))-sin(1/2)cos(1/sqrt(2))

  151. amistre64
    • 5 years ago
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    lol....right concept wrong format lol take out those sin cos stuff, its left over from my inner most deepest thoughts :)

  152. amistre64
    • 5 years ago
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    (1/sqrt(2)) (1/sqrt(3)) - (1/2) (1/sqrt(2)) 1/sqrt(6) - 1/2sqrt(2)

  153. amistre64
    • 5 years ago
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    sqrt(6)/6 - sqrt(2)/4 2sqrt(6)/12 - 3sqrt(2)/12 2sqrt(6) - 3sqrt(2) ---------------- ??? 12

  154. amistre64
    • 5 years ago
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    not it... i see where I messed it up lol

  155. amistre64
    • 5 years ago
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    (1/sqrt(2)) (sqrt(3)/2) - (1/2) (1/sqrt(2)) sqrt(6)/4 - sqrt(2)/4 sqrt(6) - sqrt(2) -------------- should be it 4

  156. anonymous
    • 5 years ago
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    You are correct :)

  157. anonymous
    • 5 years ago
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    Simplify the trigonometric expression. csc(x) - sin(x) / cot(x) Write the trigonometric expression in terms of sine and cosine, and then simplify. sin(t)sec(t)

  158. anonymous
    • 5 years ago
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    U know I'm actually learning from u :)

  159. amistre64
    • 5 years ago
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    clean that up for me....

  160. amistre64
    • 5 years ago
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    csc(x) - sin(x) / cot(x) is not very clear...

  161. anonymous
    • 5 years ago
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    Simplify the trigonometric expression. csc(x) - sin(x) over or divided by cot(x)

  162. anonymous
    • 5 years ago
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    That helps?

  163. amistre64
    • 5 years ago
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    csc(x) - sin(x) ------------ ??? cot(x)

  164. anonymous
    • 5 years ago
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    yes

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