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anonymous

  • 5 years ago

Use an addition or subtraction formula to find the exact value of the expression. sin(285)

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  1. anonymous
    • 5 years ago
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    Well, lets assume you know the 'obvious' values of sin - {0, 30, 45, 60, 90} + 360n. How can you use them to make 285?

  2. anonymous
    • 5 years ago
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    add them?

  3. anonymous
    • 5 years ago
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    Indeed. Use sin(a+b) with a couple of them and you should be able to work it out. I guess the main problem is how to do it most efficiently.

  4. anonymous
    • 5 years ago
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    so what do i add?

  5. anonymous
    • 5 years ago
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    That's where i'm cofused

  6. anonymous
    • 5 years ago
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    Something to consider is that sin (180 - x) = sin x (the best way is probably to add 2 values of sin between 100 and 180

  7. anonymous
    • 5 years ago
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    ¬_¬ Sin 150 = Sin 30 Sin 135 = Sin 45

  8. anonymous
    • 5 years ago
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    I do not get it.

  9. anonymous
    • 5 years ago
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    Do you not get why that is true?

  10. anonymous
    • 5 years ago
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    Or do you not get that sin(150 + 135) = sin 285

  11. anonymous
    • 5 years ago
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    both

  12. anonymous
    • 5 years ago
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    Well, if 0 < x < 180 , then sin (180 - x) = sin x Hence the above claim. And 135 + 150 = 285 (clearly) sin(150 + 135) = sin(150)cos(135) + sin(135)cos(150) = sin(30)cos(135) + sin(45)(cos(150) All you need to do is use a SIMILAR rule one the cos values and you're done. I'll leave that to you. NB this is how I'd do it, but there may be a quicker way.

  13. anonymous
    • 5 years ago
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    ok thanks

  14. anonymous
    • 5 years ago
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    Gah, actually I prefer: Sin 285 = sin (285 - 360) = sin(-75) Sin is an odd function => sin(-75) = -sin(75) sin 75 = sin(30+45) = blah. -sin 35 = -blah.

  15. anonymous
    • 5 years ago
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    -90?

  16. anonymous
    • 5 years ago
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    Last line is -sin(75). Where did you get 90 from?

  17. anonymous
    • 5 years ago
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    i have no idea what -sin 35= ?

  18. anonymous
    • 5 years ago
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    It's meant to be -sin(75) , sorry! But sin (75) = sin(45 + 30) = sin(45)cos(30) + sin(30)cos(45) And sin(285) = sin(-75) = -sin(75) = -(sin(45)cos(30) + sin(30)cos(45)) Because sin(-a) = -sin(a)

  19. anonymous
    • 5 years ago
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    the answer is?

  20. anonymous
    • 5 years ago
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    hmm, INew, wouldn't it be easier if you just drew the graph and compute it from the graph? :) it'll be alot easier ^_^

  21. anonymous
    • 5 years ago
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    sin(45) = cos(45) = 1/sqrt(2) sin(30) = 1/2 cos(30) = sqrt(3)/2 \[\implies -\sin(75) = -\frac{1}{\sqrt{2}} \cdot \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = -\frac{1+ \sqrt{3}}{2\sqrt{2}} = -\frac{\sqrt{2} + \sqrt{6}}{4} = \sin(285)\]

  22. anonymous
    • 5 years ago
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    Easy to compute \[-\frac{\sqrt{6} + \sqrt{2}}{4}\] from a graph? You must be pretty badass with graphs.

  23. anonymous
    • 5 years ago
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    lol, yes by drawing the graph. You know turning the lines in circles we have 90 -180 360 -270 he'll easily figure out angle by then :)

  24. anonymous
    • 5 years ago
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    any *

  25. anonymous
    • 5 years ago
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    Just admit you were wrong, please, and I won't have to own you.

  26. anonymous
    • 5 years ago
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    lol, own who?

  27. anonymous
    • 5 years ago
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    the graph is actually easier, never realized it :p

  28. anonymous
    • 5 years ago
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    LOL, do you know how yo use it loco :)?

  29. anonymous
    • 5 years ago
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    The question specifically asks to use the addition formulae, and besides, no 'circle' will give you the angle 285.

  30. anonymous
    • 5 years ago
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    lol, yes it will :) , you've never used it right?

  31. anonymous
    • 5 years ago
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    kind of, ta showed it to us. still iffy on it but practicing.

  32. anonymous
    • 5 years ago
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    besides loco, the information INew has given you also is valuable, work with both ^_^

  33. anonymous
    • 5 years ago
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    thanks guys :)

  34. anonymous
    • 5 years ago
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    np ^_^

  35. anonymous
    • 5 years ago
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    but did you figure it out ? :)

  36. anonymous
    • 5 years ago
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    sstarica, please, draw me a circle which can give the exact value I calculated without, in essence, just using the addition formulae in diagram form.

  37. anonymous
    • 5 years ago
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    I can't draw it here, but you can twirl the line till you get the given angle, then you can figure out which angles to add. Those angles will be easy to compute since they're small, and in the end you can just add them up to find the sin of them ^_^

  38. anonymous
    • 5 years ago
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    stop fighting guys

  39. anonymous
    • 5 years ago
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    Oh, adding angles? Weird, just what I did. BTW drawing is harder than saying 285 = 360 - 75 => we need -sin(75)

  40. anonymous
    • 5 years ago
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    Simplify the trigonometric expression. cos[x]/sec[x] + sin[x]/csc[x]

  41. anonymous
    • 5 years ago
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    Sorry, I thought they were implying they could do the hard part in a circle, not the easy part :/

  42. anonymous
    • 5 years ago
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    I'm not fighting LOL, we're discussing loco ^_^

  43. anonymous
    • 5 years ago
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    1/sec = cos. 1/csc = sin. cos * 1/sec = .... etc

  44. anonymous
    • 5 years ago
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    how about we i mean u guys solve this problem

  45. anonymous
    • 5 years ago
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    you still didn't solve it? .-.

  46. anonymous
    • 5 years ago
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    no, i hate this stuff

  47. anonymous
    • 5 years ago
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    don't say that, you still didn't master it , but you will with practice and it'll be easy on you ^_^

  48. anonymous
    • 5 years ago
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    i've been practicing a lot, but still having trouble :'(

  49. anonymous
    • 5 years ago
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    \[\frac{\cos x}{\sec x} = \frac{\cos x}{\frac{1}{\cos x}} = \cos^2x\]

  50. anonymous
    • 5 years ago
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    it's alright, keep on practicing and you'll master it :)

  51. anonymous
    • 5 years ago
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    sstarica help?

  52. anonymous
    • 5 years ago
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    hold on :)

  53. anonymous
    • 5 years ago
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    newton it's wrong

  54. anonymous
    • 5 years ago
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    What's wrong? (Before answering, note that I am never wrong)

  55. anonymous
    • 5 years ago
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    sorry, i thought that was the answer

  56. anonymous
    • 5 years ago
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    we all make mistakes Inew :) even I, even the most intelligent, but noone said you're wrong, but your way is complicated.

  57. anonymous
    • 5 years ago
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    try to explain it in a simple way :)

  58. anonymous
    • 5 years ago
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    I was doing one half of it. See if you can apply the same logic to the next part - you might find it simlpifies to.. something nice.

  59. anonymous
    • 5 years ago
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    He meant I am wrong about the second problem he posted in here. I only did half of it

  60. anonymous
    • 5 years ago
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    Yes, my FIRST method was comlpicated, but my second was amazing.

  61. anonymous
    • 5 years ago
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    he posted another question?

  62. anonymous
    • 5 years ago
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    lol, good that you've made it easier ^_^

  63. anonymous
    • 5 years ago
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    Thankfully still no graphs required

  64. anonymous
    • 5 years ago
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    oh that one , LOL LOCO! you've posted it in a different thread =D and I was writing it down

  65. anonymous
    • 5 years ago
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    u can write it here

  66. anonymous
    • 5 years ago
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    nvm it, INew answered it though :)

  67. anonymous
    • 5 years ago
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    it's not right though

  68. anonymous
    • 5 years ago
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  69. anonymous
    • 5 years ago
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    Alright we have :\[\cos(x)/\sec(x) + \sin(x)/\csc(x)\] sec x = 1/cosx and csc x = 1/sin x so: \[[\cos(x)/1/\cos(x)] + [\sin(x)/1/\sin(x)]\] simplify and you'll get : \[= \cos^2(x) +\sin^2(x)\] \[= 1\] Correct me if I'm wrong please ^_^

  70. anonymous
    • 5 years ago
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    you are right :) i got the same answer

  71. anonymous
    • 5 years ago
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    No, you aren't wrong. Unless you count not using \frac{}{} for fractions as wrong (I personally do)

  72. anonymous
    • 5 years ago
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    well done ^_^

  73. anonymous
    • 5 years ago
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    lol, I didn't get you INew

  74. anonymous
    • 5 years ago
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    \[\frac{1}{2} \geq 1/2\]

  75. anonymous
    • 5 years ago
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    hmm, I can't find the frac() symbol lol >_<

  76. anonymous
    • 5 years ago
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    still don't know how to make that fraction line ^_^"

  77. anonymous
    • 5 years ago
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    last problem.... Simplify the trigonometric expression. cos[x]/ tan[x] + sex[x]

  78. anonymous
    • 5 years ago
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    frac{a}{b} gives\[\frac{a}{b}\]

  79. anonymous
    • 5 years ago
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    oh, thank you :)

  80. anonymous
    • 5 years ago
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    sex(x) - lolololololol

  81. anonymous
    • 5 years ago
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    LOL

  82. anonymous
    • 5 years ago
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    sec[x], sorry :P

  83. anonymous
    • 5 years ago
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    wish math was that fun

  84. anonymous
    • 5 years ago
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    I general, always write tan as sin/cos , sec as 1/cos etc.

  85. anonymous
    • 5 years ago
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    When trying to simplify things.

  86. anonymous
    • 5 years ago
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    It's alright, we have : \[=\cos^2(x)/sinx + 1/sinx\] \[= [\cos^2x + 1]/\sin(x)\] \[= \sin^2x/sinx = \sin(x)\] ^_^

  87. anonymous
    • 5 years ago
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    ORLY?

  88. anonymous
    • 5 years ago
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    maybe? :) correct me lol

  89. anonymous
    • 5 years ago
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    it's not correct

  90. anonymous
    • 5 years ago
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    \[\cos^2x + 1 \not= \sin^2x\]

  91. anonymous
    • 5 years ago
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    hmm, where is my mistake

  92. anonymous
    • 5 years ago
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    LOL! right >_< sorry

  93. anonymous
    • 5 years ago
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    For what it's worth, I don't think it simplifies much nicer than how it starts, so is a bad question.

  94. anonymous
    • 5 years ago
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    hmm, solve it :)

  95. anonymous
    • 5 years ago
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    man, u guys are at it.

  96. anonymous
    • 5 years ago
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    LOL, loco we're not fighting =P, hold on :)

  97. anonymous
    • 5 years ago
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    Ugh, it doesn't simplify.. at all. But also sec x = 1/cos(x) ... >_> sorry

  98. anonymous
    • 5 years ago
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    \[\sec x + \csc x -\sin x\] .. but I'm tired.

  99. anonymous
    • 5 years ago
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    Loco left is for a new thread. Un coooooooooooool

  100. anonymous
    • 5 years ago
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    i posted it in a new thread also, but u guys were to busy

  101. anonymous
    • 5 years ago
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    2nd one

  102. anonymous
    • 5 years ago
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    Was my answer right?

  103. anonymous
    • 5 years ago
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    hold on

  104. anonymous
    • 5 years ago
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    Joke, I know it was. Just unsure what the 'expected' form is.

  105. anonymous
    • 5 years ago
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    it's wrong newton

  106. anonymous
    • 5 years ago
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    \[=\frac{\cos^2x}{\sin(x)} + 1/cos(x)\]\[= \frac{\cos(x)(cos^2x+1)}{\sin(x)cos(x)}\]\[= \frac{2-\sin^2x}{\sin(x)}\]\[= \frac{2}{\sin(x)} -\sin(x) = 2\csc(x) -\sin(x)\]

  107. anonymous
    • 5 years ago
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    I'm not sure though, I was laggin ^_^|"

  108. anonymous
    • 5 years ago
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    just go over it please :)

  109. anonymous
    • 5 years ago
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    nope

  110. anonymous
    • 5 years ago
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    wrong? ._.

  111. anonymous
    • 5 years ago
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    yep

  112. anonymous
    • 5 years ago
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    blah~, I'll try again ^_^

  113. anonymous
    • 5 years ago
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    For the record, loco, my answer (sec(x)+cosec(x) - sin(x)) is right, but probably thr wrong form - it's hideous.

  114. anonymous
    • 5 years ago
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    sec(x) + cos(x)cot(x), maybe?

  115. anonymous
    • 5 years ago
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    don't know, my school's hw program sucks nuts

  116. anonymous
    • 5 years ago
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    i only get 3 tries per problem

  117. anonymous
    • 5 years ago
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    That program is retarded, then. Both of the answers I've given are right, but if the form matters then both could be shown as wrong. FFFFUUUUUUUUUUUUU

  118. anonymous
    • 5 years ago
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    at least it's not timed, i've been doing this since 9pm

  119. anonymous
    • 5 years ago
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    it really pisses me off and the professor is a b**** and doesn't do anything about it

  120. anonymous
    • 5 years ago
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    lol, both of you calm down and just try again ^_^

  121. anonymous
    • 5 years ago
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    \[\frac{\cos x}{\tan x} + \sec x = \frac{\cos^2x}{\sin x} + \sec x = \cos x\cot x + \sec x = ?\]

  122. anonymous
    • 5 years ago
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    Which can also be written as sec x + csc x - sin x , and many other things.

  123. anonymous
    • 5 years ago
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    i'll wait for sstarica's answer

  124. anonymous
    • 5 years ago
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    True, she (I think, but maybe he?) does have more fans than me, so that's a good call.

  125. anonymous
    • 5 years ago
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    I'm a she =P

  126. anonymous
    • 5 years ago
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    I'm a she =P

  127. anonymous
    • 5 years ago
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    r u hot? ;)

  128. anonymous
    • 5 years ago
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    just to make sure, what is the question loco? :)

  129. anonymous
    • 5 years ago
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    let me find it

  130. anonymous
    • 5 years ago
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    Simplify the trigonometric expression. cos[x]/ tan[x] + sec[x]

  131. anonymous
    • 5 years ago
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    what on earth? lol what kind of quetion is that? >_>

  132. anonymous
    • 5 years ago
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    idk.... curious loco mind

  133. anonymous
    • 5 years ago
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    Don't worry about loco the sexpest . But in answer to the question of course she is, she does Maths.

  134. anonymous
    • 5 years ago
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    ok... lets continue with math. i don't wanna go to jail

  135. anonymous
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  136. anonymous
    • 5 years ago
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    I got : cscx -sinx + secx *sigh* not sure though, but that's what I got

  137. anonymous
    • 5 years ago
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    cos^2x + 1 ---------- sinx

  138. anonymous
    • 5 years ago
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    Blexting is about to say : This problem is trivial via the use of synthetic division.

  139. anonymous
    • 5 years ago
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    Blexting sec x is not 1 / sin x, btw.

  140. anonymous
    • 5 years ago
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    it's wrong blexting

  141. anonymous
    • 5 years ago
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    lol

  142. anonymous
    • 5 years ago
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    check mine please ^^" if I'm wrong, then I'm sorry

  143. anonymous
    • 5 years ago
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    My answer (which sstarica now has, too) has been right all along.

  144. anonymous
    • 5 years ago
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    ok this the last try i have left any final answers

  145. anonymous
    • 5 years ago
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    lol INew, you got the same answer? :)

  146. anonymous
    • 5 years ago
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    The answer =sec x + csc x - sin x =sec x + cos x cot x Now guess which is the required form :D Yes, got it above.

  147. anonymous
    • 5 years ago
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    yeah he did or she

  148. anonymous
    • 5 years ago
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    Oh, I'm a she. Hence the Newton name and everythi-- oh, wait a second ...

  149. anonymous
    • 5 years ago
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    lol, then loco go with it, I got the same answer too ^_^, damn my body temperature increased, too much concentration ^^"

  150. anonymous
    • 5 years ago
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    good question ._.

  151. anonymous
    • 5 years ago
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    nope it didn't work. sorry guys. told u the system is bs

  152. anonymous
    • 5 years ago
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    =-=......I'm sorry

  153. anonymous
    • 5 years ago
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    i got a 90 on the quiz though

  154. anonymous
    • 5 years ago
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    excellent :)

  155. anonymous
    • 5 years ago
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    Out of interest, what level is the quiz? As in, what educational level?

  156. anonymous
    • 5 years ago
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    THANKS FOR TRYING NEWTON (HE OR SHE?) AND SSTARICA (SEND ME UR # ;p)

  157. anonymous
    • 5 years ago
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    PRE-CALC

  158. anonymous
    • 5 years ago
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    lok, someone answered this question in the other thread >_< LOL!

  159. anonymous
    • 5 years ago
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    but the professor is really tough, doesn't curve or anything

  160. anonymous
    • 5 years ago
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    Oh, I'm english, so these american maths terms mean nothing to me ¬_¬ And trust me, that wasn't me trying.

  161. anonymous
    • 5 years ago
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    american ed system succks retricemate

  162. anonymous
    • 5 years ago
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    Don't worry, the person on the other thread got it wrong.

  163. anonymous
    • 5 years ago
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    bloody teachers

  164. anonymous
    • 5 years ago
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    yeah he did

  165. anonymous
    • 5 years ago
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    lol, relax, so what if the teaher sucks? depend on yourself and trust your abilities :), I had the worse teachers, so I had to depend on myself and look where I've reached :) good luck loco ^_^

  166. anonymous
    • 5 years ago
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    THANKS SSTARICA :)

  167. anonymous
    • 5 years ago
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    for my own personal interest, where have you reaced?

  168. anonymous
    • 5 years ago
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    np :)

  169. anonymous
    • 5 years ago
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    what do u mean

  170. anonymous
    • 5 years ago
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    "Look where I've reached" - I thought that meant a title/place, but now I think about I think it just means an ability level. My bad.

  171. anonymous
    • 5 years ago
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    i think it can mean both. depends on what you're talking about

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spraguer (Moderator)
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