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Homework Help: Determining Missing Trigonometric Ratios

  1. Jan 5, 2015 #1
    I'm not stuck per say, but I need to know if I have the right idea for solving the rest of these questions.

    1. The problem statement, all variables and given/known data

    For the following given trigonometric ratio and domain, determine the missing trigonometric ratio.

    2. Relevant equations

    [tex] cos\theta = -\frac{1}{2} , \frac{\pi}{2}\leq \theta < \pi [/tex] find [tex]cot\theta[/tex]

    3. The attempt at a solution

    I know how to do these normally, it's just the "find other trig. ratio" tacked on the end that's sort of distorting me. My attempt was finding the reference angle of the trig. value which turned out to be...

    [tex] \theta_{r} = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} [/tex]

    Now, the question is looking for cotangents. Because the original (cosine) was negative, I postulate that you would be looking for negative cotangents in the restriction given (not negative angles, but negative relative to the quadrant). Thus in quadrant two, the answer would be

    [tex] \pi - \frac{\pi}{3} = \frac{2\pi}{3} [/tex]

    Thank you for reading.
  2. jcsd
  3. Jan 5, 2015 #2


    Staff: Mentor

    An alternative scheme would be to recognize the triangle whose cosine is -1/2 and from there construct the cot without the need of getting the angle.
  4. Jan 5, 2015 #3


    Staff: Mentor

    No, ##\theta## is in the second quadrant. Once you know the angle in question, it's easy to get all of the trig functions of that angle.
    Yes, that's the correct angle. Now find the cotangent.
  5. Jan 5, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It would be better to solve cos-1(-1/2) for the given range before worrying about the cotangent. If cos-1(x) = A, what is the general expression for cos-1(-x)?
    Another approach is not to determine the angle at all. Just use the relationships between trig functions to go from cos to cot.

    (and the Latin expression is per se)
  6. Jan 6, 2015 #5
    Okay, I think I understand now. Thank you all who posted.
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