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Check please

  1. Jan 11, 2005 #1
    [tex] \cos \theta = \frac {2} {3} and \frac {3\pi} {2} <\theta<2\pi [/tex] then determine the exact value of [tex] \frac {1} {\cot\theta} [/tex]

    I did this question using sin^2+cos^2=1 subbing in cos and then solving for sin when i got both values i realized that 1/cot theta = tan theta which is (sin theta)/(cos theta)

    I got my answer to be squareroot 5/2 is this correct?
  2. jcsd
  3. Jan 11, 2005 #2
    Looks good.
  4. Jan 11, 2005 #3
    I think it is correct if what you mean is [tex]\frac{\sqrt5}{2}[/tex], and not [tex]\sqrt\frac{5}{2}[/tex].
  5. Jan 11, 2005 #4
    yes the first one thanks
  6. Jan 11, 2005 #5
    Actually, I'm starting to have second thoughts about this. If you used a calculator to work out [tex]\theta[/tex] using the initial information given, and then substitute this value into solving [tex]\tan(\theta)[/tex], the answer you get is quite different. Hmm...
  7. Jan 11, 2005 #6
    little aisha,
    why don't you use calculator to check your problems? I am sure this is faster and easier for you... I knew you are not allowed to use calculator DO the problems, but for checking purpose only. I think your teacher could forgive you. I am not saying you shouldn't post your hw here, but since you got the answer already... I think you are wasting your time typing every HW here just to check the answer while you can do it in an faster way. by the way, if you have any question, you are welcome and we are glad to help you....
  8. Jan 11, 2005 #7
    its ok i dont mind wasting time, its an online course and most of the question i post are for assignments which dont have solutions, instead of checking myself i want to be fully satisfied that my answer is right, sorry if that bothers u
  9. Jan 11, 2005 #8
    I think you might have punched it in wrong. The answer is right..
  10. Jan 11, 2005 #9
    Well, lets see.

    [tex]\theta = \cos^{-1}\frac{2}{3} = 48.18968511 or 311.8103149[/tex]

    However, it has already been stated that [tex]\frac {3\pi} {2} <\theta<2\pi [/tex]. So [tex] \theta = 311.8103149[/tex].

    [tex]\frac {1} {\cot\theta} = \tan\theta = \tan 311.8103149 = -1.118033989 [/tex]

    Can someone please tell me what is wrong with this?
    Last edited: Jan 11, 2005
  11. Jan 11, 2005 #10


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    So obviously the answer has to be [tex]\tan \theta = -\frac{\sqrt{5}}{2}[/tex]
  12. Jan 11, 2005 #11
    Oh dear, Aisha is missing a negative sign. Well, that's what you get for not using the calculator...
  13. Jan 11, 2005 #12


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    The confusion is not because of using a calculator, it's because of lack of understanding of the basics. Aisha, you have to remember in which quadrants each trig ratio is positive and negative. I remember it as "a, s, t, c". I had a mnemonic a long time ago, but discarded it after it became second nature. a = "ALL" positive (1st quadrant), s = "SIN" positive (2nd quadrant), t = "TAN" positive (3rd quadrant) and c = "COS" positive (4th quadrant).

    Here a 4th quadrant answer is expected so the cosine will be positive while the tangent (and cotangent) will be negative.
  14. Jan 11, 2005 #13
    What I meant was that she should have used a calculator for double checking.
  15. Jan 11, 2005 #14


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    Calculators can become a crutch. Think of the danger here...she uses her answer of the positive root and punches it in, extracting an arctangent and gets a principal value in the first quadrant. But it doesn't hit her that it's a first quadrant answer so she blithely takes the cosine of it and the answer comes out to 2/3. She then submits the wrong answer.

    See the problem ? Calculators are for improving the efficiency of people who already know what they're doing, not for correcting the errors of people who don't yet know.
  16. Jan 11, 2005 #15
    Well, I quite agree with what you're saying about people becoming overly dependant on calculators. It is also true that she is having some difficulties with understanding quadrants (as demonstrated in one of her previous threads).
  17. Jan 11, 2005 #16


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    Fully agree. :smile:
  18. Jan 12, 2005 #17
    lol wow i understand the quadrants and what is positive and negative
    I know C A S T for each quadrant

    I just forgot that sin is negative in the 4th quadrant thats why I made a little error, but thanks for all ur help in checking this question I really appreciate it :smile:
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