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kittiwitti1

  • one year ago

Problems which I cannot for the life of me figure out why I got wrong/partially incorrect: A: Find all six trigonometric functions of θ if the given point is on the terminal side of θ. (Assume that a is a positive number. If an answer is undefined, enter UNDEFINED.) (−5a, −12a) Result: http://prntscr.com/8diac1 B: Use your calculator to find sin θ and cos θ if the point (9.37, 7.05) is on the terminal side of θ. (Round your answers to four decimal places.) Result: http://prntscr.com/8dia9h C: Indicate the quadrants in which the terminal side of θ must lie under the following conditions. (Select all that apply.) cos θ and cot θ have the same sign Result: http://prntscr.com/8diaew

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  1. anonymous
    • one year ago
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    |dw:1441605814325:dw| So we get \[\sin(\theta)=\sin(180+x)=-\sin(x)\] \[\cos(\theta)=\cos(180+x)=-\cos(x)\]\[\tan(\theta)=\tan(180+x)=\tan(x)\] from the triangle we can get \[\sin(x)=\frac{-12a}{\sqrt{(-12a)^2+(-5a)^2}}=\frac{-12a}{\sqrt{144a^2+25a^2}}=\frac{-12a}{13a}=-\frac{12}{13}\] Similarly find cos(x) and tan(x) and use them to find sin(theta), cos(theta) and tan(theta) reciprocal for cosec(theta) sec(theta) and cot(theta)

  2. kittiwitti1
    • one year ago
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    Which problem are you doing? (Sorry I'm multitasking) e_e

  3. anonymous
    • one year ago
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    Do you have a figure?Talking about problem 1

  4. kittiwitti1
    • one year ago
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    I'm only wondering where I went wrong, I don't need the whole process explained e_e;;;

  5. kittiwitti1
    • one year ago
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    Besides they said the sin was correct... My issue is how the tan/cot are incorrect.

  6. anonymous
    • one year ago
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    I see your problem now, what you did was, you found the ratios \[\sin(x),\cos(x),\tan(x)\] BUT you're suppose to find \[\sin(\theta),\cos(\theta),\tan(\theta)\] and NOT \[\sin(x),\cos(x),\tan(x)\] what I mean by x, refer to the figure

  7. kittiwitti1
    • one year ago
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    I don't see what you mean by x... yes I looked at the figure. x_x;

  8. anonymous
    • one year ago
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    You did something like |dw:1441606633005:dw|

  9. kittiwitti1
    • one year ago
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    Yes but aren't the trig values the same?

  10. anonymous
    • one year ago
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    Nope!! It makes a slight difference

  11. kittiwitti1
    • one year ago
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    It does? *trig function values

  12. anonymous
    • one year ago
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    Refer to these formulae I've given below \[\sin(\theta)=\sin(180+x)=-\sin(x)\]\[\cos(\theta)=\cos(180+x)=-\cos(x)\]\[\tan(\theta)=\tan(180+x)=\tan(x)\]

  13. anonymous
    • one year ago
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    As you can see sign changes for sine and cosine but remains same for tangent

  14. anonymous
    • one year ago
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    Let's find your tan theta, since it's seem to be giving you problem \[\tan(\theta)=\tan(x)=\frac{-12a}{-5a}=\frac{12}{5}\]

  15. kittiwitti1
    • one year ago
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    Er. Yes...

  16. anonymous
    • one year ago
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    Actually you've mixed your adjacent and opposite for tangent

  17. anonymous
    • one year ago
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    that's why you're getting wrong answer

  18. anonymous
    • one year ago
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    and if your tangent is wrong then obviously cotangent will also be wrong

  19. anonymous
    • one year ago
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    notice you answers you've done \[\frac{5}{12}\] when it should be \[\frac{12}{5}\]

  20. anonymous
    • one year ago
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    @kittiwitti1 you there ??

  21. kittiwitti1
    • one year ago
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    Sorry, working on other problems... What'd I miss?

  22. anonymous
    • one year ago
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    for problem 1 actually you've exchanged your adjacent and opposite accidentally

  23. kittiwitti1
    • one year ago
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    -facedesks- Dx I feel dumb. Okay, thanks; what about the other two?

  24. anonymous
    • one year ago
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    problem 3) Quadrant 2, both cos and cot have negative sign so that's ok Quadrant 3, cos is negative while cot is POSITIVE, so that's wrong how about Quadrant 1?both cos and cot, infact all functions are positive!

  25. kittiwitti1
    • one year ago
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    B: http://prntscr.com/8dia9h C: http://prntscr.com/8diaew

  26. anonymous
    • one year ago
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    |dw:1441607496705:dw|

  27. freckles
    • one year ago
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    \[\sin(\theta)=\frac{y}{r} \text{ and } \cos(\theta)=\frac{x}{r} \\ \text{ and you have } r=\sqrt{x^2+y^2} \\ \text{ you can use this \to help you evaluate } \sin(\theta) \text{ and } \cos(\theta)\]

  28. kittiwitti1
    • one year ago
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    Oh... I seem to have attracted quite the audience... sorry for previous "mass-tags" ^^;

  29. anonymous
    • one year ago
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    |dw:1441607639710:dw|

  30. kittiwitti1
    • one year ago
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    Wait what are you going on about lol

  31. anonymous
    • one year ago
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    After->1st quadrant School->2nd quadrant To->3rd quadrant College->4th quadrant

  32. anonymous
    • one year ago
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    It's an easy way to remember

  33. anonymous
    • one year ago
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    like soh cah toa

  34. kittiwitti1
    • one year ago
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    Yes but what does that have to do with problem B o_o (I mean thanks for the info but I don't see the connection)

  35. anonymous
    • one year ago
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    In problem 3 quadrant is wrong look at it again

  36. anonymous
    • one year ago
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    3rd quadrant

  37. anonymous
    • one year ago
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    is wrong

  38. anonymous
    • one year ago
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    I told you about that just in case you have any trouble remembering what is positive in which quadrant

  39. kittiwitti1
    • one year ago
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    Oh okay that was the answer for problem 3, but what about #2 o_o?

  40. anonymous
    • one year ago
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    No, I'm saying that your choice of quadrant 3 is wrong, your answer should be quadrant 2 and 1 other quadrant, think about it

  41. kittiwitti1
    • one year ago
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    I was looking at my notes... did the teacher teach us the wrong things D:

  42. kittiwitti1
    • one year ago
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    Is it Q1-Q3?

  43. freckles
    • one year ago
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    cosine and cotangent have the same sign that means they could both be positive or they could both be negative if cosine and cotangent are both negative then what is sine going to be? if cosine and cotangent are both positive then what is sine going to be?

  44. kittiwitti1
    • one year ago
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    So Q1-Q3?

  45. kittiwitti1
    • one year ago
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    Okay

  46. anonymous
    • one year ago
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    Q3 is not the correct option as I said earlier

  47. kittiwitti1
    • one year ago
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    Wait what?

  48. anonymous
    • one year ago
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    in Q3 cos is negative while cot is positive not the same sign at all

  49. kittiwitti1
    • one year ago
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    My notes say Q3 has both negative values for cos and cot.

  50. freckles
    • one year ago
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    cosine and cotangent have the same sign that means they could both be positive or they could both be negative if cosine and cotangent are both negative then what is sine going to be? then sine is positive if cosine and cotangent are both positive then what is sine going to be? then sine is positive yes he is right!

  51. freckles
    • one year ago
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    @Nishant_Garg is right

  52. freckles
    • one year ago
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    I should have answered my own questions

  53. kittiwitti1
    • one year ago
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    I just realized I wrote a wrong trig function in my notes...

  54. kittiwitti1
    • one year ago
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    the* wrong Nevermind. What about problem B?

  55. anonymous
    • one year ago
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    you've written wrong dear. Take it from me, never refer to notes for absolute answers because we students tend to write notes during a class while trying to fight off sleep, not very accurate for your notes huh? :P

  56. kittiwitti1
    • one year ago
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    Lol but I already noticed before you'd said that. Actually the teacher had miswritten. I should have read the book... forgot the teacher always makes mistakes during lectures (she took 8 minutes till she realized she was going over the wrong problem once)

  57. kittiwitti1
    • one year ago
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    Even though we tried to tell her five times... What about problem B? http://prntscr.com/8dia9h

  58. anonymous
    • one year ago
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    Ouch, for problem 2 you can do the following remember the unit circle? we have the following equations \[x=r\cos(\theta), y=r\sin(\theta)\] Dividing these we get \[\frac{y}{x}=\frac{r\sin(\theta)}{r\cos(\theta)}=\tan(\theta)\] \[\therefore \tan(\theta)=\frac{y}{x}=\frac{7.05}{9.37}\] Now that you've found 1 trignometric ratio, you can find any other using identities \[1+\tan^2(\theta)=\sec^2(\theta) \implies \sec(\theta)=\sqrt{1+\tan^2(\theta)} \implies \cos(\theta)=\frac{1}{\sqrt{1+\tan^2(\theta)}}\]

  59. kittiwitti1
    • one year ago
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    http://prntscr.com/8dilgn Actually I need help on this first

  60. kittiwitti1
    • one year ago
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    LOL sorry

  61. kittiwitti1
    • one year ago
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    Mostly because idk what that symbol means

  62. anonymous
    • one year ago
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    So which one first?

  63. kittiwitti1
    • one year ago
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    The one posted just now

  64. kittiwitti1
    • one year ago
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    I only need to know what the symbol is really

  65. anonymous
    • one year ago
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    ok \[\cos(\theta)=\frac{2\sqrt{13}}{13}\] One ratio you can find right off the start is secant \[\sec(\theta)=\frac{1}{\cos(\theta)}=\frac{13}{2\sqrt{13}}\]

  66. anonymous
    • one year ago
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    \[\in\] This symbol?

  67. kittiwitti1
    • one year ago
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    Yes, that one. D:

  68. anonymous
    • one year ago
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    It means "belongs to" So what they're saying is theta belongs to quadrant IV You'll learn more about the symbol as you'll read set theory

  69. anonymous
    • one year ago
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    what grade are you in?

  70. kittiwitti1
    • one year ago
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    1st? o_o (College)

  71. anonymous
    • one year ago
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    The reason they've given this information will be clear after you read this suppose we want to find sin(theta) now we can do \[\sin^2(\theta)+\cos^2(\theta)=1\]\[\sin^2(\theta)=1-\cos^2(\theta)\]\[\implies |\sin(\theta)|=\sqrt{1-\cos^2(\theta)}\]\[\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}\] Since theta is in 4th quadrant, sign will be negative for sin(Theta) so we have \[\sin(\theta)=-\sqrt{1-\cos^2(\theta)}\]

  72. anonymous
    • one year ago
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    How come you're studying such basic trigonometry in college? O.o

  73. kittiwitti1
    • one year ago
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    Didn't take trig in high school. ...the timer reset at 11:59 and erased half the questions so I got 51%... WTF internet

  74. anonymous
    • one year ago
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    Oh, our syllabus is fixed for school, every student that studies mathematics in high school has studied trigonometry and basic calculus here

  75. kittiwitti1
    • one year ago
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    http://prntscr.com/8dipej ...I hate this website. Hence why I prefer PHYSICAL tests!

  76. anonymous
    • one year ago
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    >_<

  77. kittiwitti1
    • one year ago
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    Well since I didn't finish on time apparently (aka stupid website cleared off half my answers) I guess I'll just close the question = =;

  78. anonymous
    • one year ago
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    sad :/

  79. kittiwitti1
    • one year ago
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    Yeah :( oh well.

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