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Basic trig test question

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Given that cot [itex]\theta[/itex] = -12/5 and csc [itex]\theta[/itex] < 0, find sec[itex]\theta[/itex].

    This was a question on a test that I drew a blank on, and I'm still not sure how to handle it due to my "teacher" repeatedly dismissing me when I try asking about it. Now, it occurred to me that this could be solved using trig identities and substitution. Starting with [itex]sin^{2}[/itex][itex]\theta[/itex] + [itex]cos^{2}[/itex][itex]\theta[/itex] = 1 and sin/cos = -5/12, I ended up with -
    sin = -5/13
    cos= 12/13
    tan= -5/12
    sec= 13/12

    and I am confident this is the right answer. But, we have not covered trig identities in class so I am sure there is another easier way to solve this. My question is...what is it? Also, say you take the square root of a trig identity in an equation - how do you know weather it is positive or negative? Thanks in advance.
     
    Last edited: Mar 9, 2012
  2. jcsd
  3. Mar 9, 2012 #2

    tiny-tim

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    Homework Helper

    hi e^(i Pi)+1=0! :smile:
    that is the way :smile:

    (though it would be easier to memorise and use sec2 = tan2 + 1, csc2 = cot2 + 1 :wink:)

    (another way of course is to say that if cot = 12/5, then it's obviously a 5,12,13 triangle, and then eg sec will be hyp/adj)
    you need to be told (as in this question)

    btw, i can't see any latex :redface: … are other people having this problem?​
     
  4. Mar 9, 2012 #3
    Even if you haven't covered [itex]sin^{2}[/itex][itex]\theta[/itex] + [itex]cos^{2}[/itex][itex]\theta[/itex] = 1 formally, I guess you could envisage a right-angled triangle with adjacent 12 and opposite 5, and get the hypotenuse with Pythagoras, for one.
     
  5. Mar 9, 2012 #4
    Sometimes I don't see latex, but it always pops up after I refresh. Thank you for the quick responses.

    edit - actually, my answer WAS wrong since I started with tan = -12/5 when it was -5/12, but it's fixed now.
     
    Last edited: Mar 9, 2012
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